Jules Louis Philippe Coignet (1798-1860) was a French painter.
We have sections
\begin{equation}{\label{a}}\tag{A}\mbox{}\end{equation}
The Derivative.
Definition. A function \(f\) is said to be differentiable at \(x\) when the following limit
\[\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}\]
exists. If this limit exists, it is called the derivative of \(f\) at \(x\), and is denoted by \(f'(x)\). The function \(f\) is a differentiable function if it is differentiable at each \(x\) in its domain. \(\sharp\)
Geometrically, we see that the derivative
\[f'(x)=\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}\]
is the slope of the graph at the point \((x,f(x))\). The line that passes through the point \((x,f(x))\) with slope \(f'(x)\) is called the tangent line to the graph of \(f\) at \((x,f(x))\). This is the line that best approximates the graph of \(f\) near the point \((x,f(x))\). We begin with a function \(f\) and choose a point \((x_{0},y_{0})\) on the graph. If \(f\) is differentiable at \(x_{0}\), then the tangent line through this point has slope \(f'(x_{0})\). To get an equation for this tangent line, we use the point-slope formula
\[y-y_{0}=m(x-x_{0}).\]
In this case \(m=f'(x_{0})\) and the equation becomes
\[y-y_{0}=f'(x_{0})(x-x_{0}).\]
Definition. The left-hand derivative of \(f\) at \(x\), denoted by \(f_{-}^{\prime}(x)\), is defined by
\[f_{-}^{\prime}(x)=\lim_{h\rightarrow 0-}\frac{f(x+h)-f(x)}{h}.\]
The right-hand derivative of \(f\) at \(x\), denoted by$f_{+}^{\prime}(x)$, is defined by
\[f_{+}^{\prime}(x)=\lim_{h\rightarrow 0+}\frac{f(x+h)-f(x)}{h}.\]
It is clear to see that \(f(x)\) is differentiable at \(x\) if and only if
\[f'(x)=f’_{-}(x)=f’_{+}(x).\]
Setting \(x=c+h\), we can write
\[f'(c)=\lim_{x\rightarrow c}\frac{f(x)-f(c)}{x-c}.\]
Example. Given a function
\[f(x)=\left\{\begin{array}{ll}
x^{2}, & x\leq 1\\
2x-1, & x>1
\end{array}\right .,\]
Find \(f'(1)\). By definition, we have
\[f'(1)=\lim_{h\rightarrow 0} \frac{f(1+h)-f(1)}{h}.\]
Therefore, we obtain
\begin{align*}
\lim_{h\rightarrow 0-}\frac{f(1+h)-f(1)}{h} & =\lim_{h\rightarrow 0-}\frac{(1+h)^{2}-1}{h}\\
& =\lim_{h\rightarrow 0-} (2+h) = 2.
\end{align*}
and
\begin{align*}
\lim_{h\rightarrow 0+}\frac{f(1+h)-f(1)}{h} & =\lim_{h\rightarrow 0+}\frac{[2(1+h)-1]-1}{h}\\
& =\lim_{h\rightarrow 0+} 2=2.
\end{align*}
We conclude \(f'(1)=2\). \(\sharp\)
Theorem. Suppose that \(f\) is differentiable at \(x\). Then \(f\) is continuous at \(x\). \(\sharp\)
\begin{equation}{\label{b}}\tag{B}\mbox{}\end{equation}
Some Differentiation Formulas.
Theorem. Let \(\alpha\) be a real number. Suppose that \(f\) and \(g\) are differentiable at \(x\). Then \(f+g\) and \(\alpha f\) are differentiable at \(x\). Moreover, we have
\[(f+g)'(x)=f'(x)+g'(x)\]
and
\[(\alpha f)'(x)=\alpha f'(x).\]
Since \(f-g=f+(-1)g\), it follows that if \(f\) and \(g\) are differentiable at \(x\), then \(f-g\) is differentiable at \(x\) with
\[(f-g)'(x)=f'(x)-g'(x).\]
Suppose that \(f_{1},f_{2},\cdots ,f_{n}\) are differentiable at \(x\). Then, given any real numbers \(\alpha_{1},\alpha_{2},\cdots ,\alpha_{n}\), the function
\[\alpha_{1}f_{1}+\alpha_{2}f_{2}+\cdots +\alpha_{n}f_{n}\]
is differentiable at \(x\) with
\[(\alpha_{1}f_{1}+\alpha_{2}f_{2}+\cdots +\alpha_{n}f_{n})'(x)=\alpha_{1}f’_{1}(x)+\alpha_{2}f’_{2}(x)+\cdots +\alpha_{n}f’_{n}(x).\]
Theorem. (Product Rule). Suppose that \(f\) and \(g\) are differentiable at \(x\). Then, the product \(fg\) is also differentiable at \(x\) with
\[(fg)'(x)=f(x)g'(x)+g(x)f'(x).\]
Using the product rule, it is not difficult to prove that, for each positive integer \(n\), the polynomial \(p(x)=x^{n}\) has derivative \(p'(x)=nx^{n-1}\) by using mathematical induction on \(n\). The proof is given below.
- (Initial condition). This is true for \(n=1\).
- (Induction hypothesis). Suppose that \(n=k\) is true.
- We are going to prove that \(n=k+1\) is true. Now, we have \(p(x)=x^{k+1}=x^{k}\cdot x\). Therefore, we obtain
\begin{align*}
p'(x) & =(x^{k})^{\prime}\cdot x+x^{k}\cdot 1 \mbox{ (using the product rule and initial condition)}\\
& =kx^{k-1}\cdot x+x^{k}\mbox{ (using the induction hypothesis)}\\
& =(k+1)x^{k}.
\end{align*}
In general, given a polynomial
\[p(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0},\]
its derivative is given by
\[p'(x)=na_{n}x^{n}+(n-1)a_{n-1}x^{n-1}+\cdots +2a_{2}x+a_{1}.\]
Example. Differentiate
\[F(x)=(x^{3}-2x+3)(4x^{2}+1).\]
We have a product \(F(x)=f(x)g(x)\) with \(f(x)=x^{3}-2x+3\) and \(g(x)=4x^{2}+1\). The product rule gives
\begin{align*}
F'(x) & =f'(x)g(x)+f(x)g'(x)\\
& =20x^{4}-21x^{2}+24x-2.
\end{align*}
Theorem. Suppose that \(g\) is differentiable at \(x\) with \(g(x)\neq 0\). Then \(1/g\) is differentiable at \(x\) with
\[\left (\frac{1}{g}\right )'(x)=-\frac{g'(x)}{[g(x)]^{2}}.\]
Let \(n\) be a negative integer. The function \(p(x)=x^{n}=1/x^{-n}\) has derivative
\[p'(x)=-\frac{(x^{-n})’}{x^{-2n}}=n\cdot\frac{x^{-n-1}}{x^{-2n}}=nx^{n-1}.\]
Therefore, for any integer \(n\), the function \(p(x)=x^{n}\) has derivative \(p'(x)=nx^{n-1}\).
Example. Differentiate
\[f(x)=\frac{1}{ax^{2}+bx+c}.\]
Let \(g(x)=ax^{2}+bx+c\). Then, we have
\[f'(x)=-\frac{g'(x)}{[g(x)]^{2}}=-\frac{2ax+b}{(ax^{2}+bx+c)^{2}}.\]
Theorem. (Quotient Rule). Suppose that \(f\) and \(g\) are differentiable at \(x\) with \(g(x)\neq 0\). Then, the quotient \(f/g\) is differentiable at \(x\) with
\[\left (\frac{f}{g}\right )'(x)=\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^{2}}.\]
Example. Differentiate
\[f(x)=\frac{ax+b}{cx+d}.\]
We have
\[f'(x)=\frac{(cx+d)a-(ax+b)c}{(cx+d)^{2}}=\frac{ad-bc}{(cx+d)^{2}}.\]
Theorem. (Chain Rule). Suppose that \(g\) is differentiable at \(x\), and that \(f\) is differentiable at \(g(x)\). Then, the composition function \(f\circ g(x)=f(g(x))\) is differentiable at \(x\) with
\[(f\circ g)'(x)=f'(g(x))g'(x).\]
Example. Differentiate
\[F(x)=\left (x+\frac{1}{x}\right )^{-3}.\]
Let \(g(x)=x+\frac{1}{x}\) and \(f(x)=x^{-3}\). Then \(F(x)=f(g(x))\). Therefore, we have
\[F'(x)=-3\left (x+\frac{1}{x}\right )^{-4}\cdot g'(x)=-3\left (x+\frac{1}{x}\right )^{-4}\left (1-\frac{1}{x^{2}}\right ).\]
Using the Leibniz notation, we write
\[f'(x)=\frac{d}{dx}[f(x)].\]
If \(y=f(x)\), we also write
\[f'(x)=\frac{dy}{dx}.\]
Then, we have
\begin{align*}
\frac{d}{dx}[f(x)+g(x)] & =\frac{d}{dx}[f(x)]+\frac{d}{dx}[g(x)]\\
\frac{d}{dx}[f(x)-g(x)] & =\frac{d}{dx}[f(x)]-\frac{d}{dx}[g(x)]\\
\frac{d}{dx}[\alpha f(x)] & =\alpha\frac{d}{dx}[f(x)]\\
\frac{d}{dx}[f(x)g(x)] & =f(x)\frac{d}{dx}[g(x)]+g(x)\frac{d}{dx}[f(x)]\\
\frac{d}{dx}\left [\frac{1}{g(x)}\right ] & =-\frac{1}{[g(x)]^{2}}\frac{d}{dx}[g(x)]\\
\frac{d}{dx}\left [\frac{f(x)}{g(x)}\right ] & =\frac{{\displaystyle g(x)\frac{d}{dx}[f(x)]-f(x)\frac{d}{dx}[g(x)]}}{[g(x)]^{2}}.
\end{align*}
Often functions \(f\) and \(g\) are replaced by \(u\) and \(v\) and the \(x\) is left out altogether. Then, the formulas look like
\begin{align*}
\frac{d}{dx}(u+v) & =\frac{du}{dx}+\frac{dv}{dx}\\
\frac{d}{dx}(u-v) & =\frac{du}{dx}-\frac{dv}{dx}\\
\frac{d}{dx}(\alpha u) & =\alpha\frac{du}{dx}\\
\frac{d}{dx}(uv) & =u\frac{dv}{dx}+v\frac{du}{dx}\\
\frac{d}{dx}\left (\frac{1}{v}\right ) & =-\frac{1}{v^{2}}\frac{dv}{dx}\\
\frac{d}{dx}\left (\frac{u}{v}\right ) & =\frac{{\displaystylev\frac{du}{dx}-u\frac{dv}{dx}}}{v^{2}}.
\end{align*}
Suppose that \(y\) is a differentiable function of \(u\) given by \(y=f(u)\), and that \(u\) is a differentiable function of \(x\) given by \(u=g(x)\). Then \(y\) is a composition function of \(x\) given by \(y=f(u)=f(g(x))\). Then, the chain rule theorem can be written by
\[\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}\mbox{ or }\frac{d}{dx}[f(u)]=\frac{d}{du}[f(u)]\frac{du}{dx}.\]
Example. Differentiate
\[f(x)=[1-(2+3x)^{2}]^{3}.\]
We have
\begin{align*} f'(x) & =\frac{d}{dx}[1-(2+3x)^{2}]^{3}\\ & =3[1-(2+3x)^{2}]^{2}\cdot\frac{d}{dx}[1-(2+3x)^{2}].\end{align*}
Since
\begin{align*} \frac{d}{dx}[1-(2+3x)^{2}] & =-2(2+3x)\cdot\frac{d}{dx}(2+3x)\\ & =-2(2+3x)3=-6(2+3x),\end{align*}
it follows
\begin{align*} f'(x) & =\frac{d}{dx}[1-(2+3x)^{2}]^{3}\\ & =3[1-(2+3x)^{2}]^{2}\cdot [-6(2+3x)]\\ & =-18(2+3x)[1-(2+3x)^{2}]^{2}.\end{align*}
Example. Differentiate
\[f'(x)=4x(x^{2}+3)^{3}.\]
We have
\begin{align*}
\frac{d}{dx}[4x(x^{2}+3)^{3}] & =4x\cdot\frac{d}{dx}[(x^{2}+3)^{3}]+(x^{2}+3)^{3}\cdot\frac{d}{dx}(4x)\\
& =4x[3(x^{2}+3)^{2}\cdot\frac{d}{dx}(x^{2}+3)]+4(x^{2}+3)^{3}\\
& =4x[3(x^{2}+3)^{2}\cdot 2x]+4(x^{2}+3)^{3}\\
& =4(x^{2}+3)^{2}(7x^{2}+3).
\end{align*}
Differentiating the Trigonometric Functions.
We are going to consider the differentiation of the trigonometric functions. Their differentiations are summarized in the following table.
\[\begin{array}{|c|c|}\hline
f(x) & f'(x)\\ \hline
\sin x & \cos x\\ \hline
\cos x & -\sin x\\ \hline
\tan x & \sec^{2} x\\ \hline
\cot x & -\csc^{2} x\\ \hline
\sec x & \sec x\tan x\\ \hline
\csc x & -\csc x\cot x\\ \hline
\end{array}\]
Example. Differentiating the following function
\[f(x)=\frac{1-\sec x}{\tan x}.\]
We have
\begin{align*}
\frac{d}{dx}\left [\frac{1-\sec x}{\tan x}\right ] & =\frac{{\displaystyle \tan x\frac{d}{dx}(1-\sec x)-(1-\sec x)\frac{d}{dx}(\tan x)}}{\tan^{2} x}\\
& =\frac{\tan x(-\sec x\tan x)-(1-\sec x)(\sec^{2} x)}{\tan ^{2} x}\\
& =\frac{\sec x(sec^{2} x-\tan^{2} x)-\sec^{2} x}{\tan^{2} x}\\
& =\frac{\sec x-\sec^{2} x}{\tan^{2} x}
\end{align*}
Example. Differentiating the following function
\[f(x)=\sec (x^{2}+1).\]
We have
\begin{align*}
\frac{d}{dx}[\sec (x^{2}+1)] & =\sec (x^{2}+1)\tan (x^{2}+1)\frac{d}{dx}(x^{2}+1)\\
& =2x\sec (x^{2}+1)\tan (x^{2}+1).
\end{align*}
Example. Differentiating the following function
\[f(x)=\sin^{3} \pi x.\]
We have
\begin{align*}
\frac{d}{dx}(\sin^{3} \pi x) & =\frac{d}{dx} (\sin \pi x)^{3}\\
& =3(\sin \pi x)^{2}\frac{d}{dx}(\sin \pi x)\\
& =3(\sin \pi x)^{2}\cos \pi x\frac{d}{dx}(\pi )\\
& =3\pi\sin^{2} \pi x\cos \pi x\end{align*}
\begin{equation}{\label{c}}\tag{C}\mbox{}\end{equation}
Implicit Differentiation.
The function that we have been studying up to this point have been defined by expressing the dependent variable explicitly in terms of the independent variable. For example, \(y=f(x)\) or \(z=g(y)\). Now, we are going to consider functional relationships between two variables in which the dependent variable is not given explicitly in terms of the independent variable. For example, suppose we know that \(y\) is a function of \(x\) that satisfies the equation
\[3x^{3}y-4y-2x+1=0.\]
We say that this equation defines \(y\) implicitly as a function of \(x\). One way to find \(dy/dx\) is to solve first for \(y\)
\[y=\frac{2x-1}{3x^{3}-4}.\]
and then differentiate
\[\frac{dy}{dx}=-\frac{12x^{3}-9x^{2}+8}{(3x^{3}-4)^{2}}.\]
It is also possible to find \(dy/dx\) without solving the equation for \(y\). The technique is called implicit differentiation. We differentiate both sides of the equation with respect to \(x\). Then, we have
\[\frac{d}{dx}(3x^{3}y-4y-2x+1)=\frac{d}{dx}(0),\]
which implies
\[\left (3x^{3}\frac{dy}{dx}+9x^{2}y\right )-4\frac{dy}{dx}-2=0.\]
Therefore, we obtain
\[\frac{dy}{dx}=\frac{2-9x^{2}y}{3x^{3}-4}.\]
The answer looks different from what we obtained before because this time \(y\) appears on the right-hand side. The two answer are really the same if we substitute
\[y=\frac{2x-1}{3x^{3}-4}\]
into the above expression. Implicit differentiation is particularly useful where it is inconvenient (or impossible) to first solve the given equation for \(y\).
Example. Use implicit differentiation to find \(dy/dx\).
\[2xy-y^{3}+1=x+2y.\]
We have
\[\left (2x\frac{dy}{dx}+2y\right )-3y^{2}\frac{dy}{dx}=1+2\frac{dy}{dx},\]
which implies
\[\frac{dy}{dx}=\frac{1-2y}{2x-3y^{2}-2}.\]
Example. Use implicit differentiation to find \(dy/dx\).
\[\cos (x-y)=(2x+1)^{2}y.\]
We have
\[-\sin (x-y)\left [1-\frac{dy}{dx}\right ]=(2x+1)^{2}\frac{dy}{dx}+2(x+1)(2)y,\]
which impies
\[frac{dy}{dx}=\frac{4(2x+1)y+\sin (x-y)}{\sin (x-y)-(2x+1)^{2}}.\]
Example. Find the slope of the tangent line of the curve \(x^{3}+y^{3}=1+3xy^{2}\) at point \((2,-1)\). Using implicit differentiation to find \(dy/dx\), we have
\[\frac{d}{dx}(x^{3}+y^{3})=\frac{d}{dx}(1+3xy^{2}),\]
which implies
\[3x^{2}+3y^{2}\frac{dy}{dx}=3x\left (2y\frac{dy}{dx}\right ).\]
We solve
\[\frac{dy}{dx}=\frac{y^{2}-x^{2}}{y^{2}-2xy}\]
At \(x=2\) and \(y=1\), we find that \(dy/dx=-3/5\). The slope is \(-3/5\). \(\sharp\)
We have shown that the differentiation formula
\[\frac{d}{dx}(x^{n})=nx^{n-1}.\]
holds for all integers \(n\). We can also show that this formula holds for rational powers \(p/q\); that is, we have
\[\frac{d}{dx}(x^{\frac{p}{q}})=\frac{p}{q}x^{\frac{p}{q}-1}.\]
The proof is as follows. Let \(y=x^{p/q}\), where \(p\) and \(q\) are nonzero integers. Then \(y^{q}=x^{p}\). Using the implicit differentiation, we have
\[qy^{q-1}\frac{dy}{dx}=px^{p-1}\]
and
\[\frac{dy}{dx}=\frac{p}{q}x^{p-1}y^{1-q}=\frac{p}{q}x^{p-1}(x^{p/q})^{1-q}=\frac{p}{q}x^{p/q-1}.\]
Some examples are given below
\begin{align*}
& \frac{d}{dx}(\sqrt[3]{x^{2}})=\frac{d}{dx}(x^{\frac{2}{3}})=\frac{2}{3}x^{-\frac{1}{3}}\\
& \frac{d}{dx}(\sqrt{x^{5}})=\frac{d}{dx}(x^{\frac{5}{2}})=\frac{5}{2}x^{\frac{3}{3}}\\
& \frac{d}{dx}\left (\frac{1}{\sqrt[9]{x^{7}}}\right )=\frac{d}{dx}x^{-\frac{7}{9}}=-\frac{7}{9}x^{-\frac{16}{9}}.
\end{align*}
Example. Differentiating the following function
\[f(x)=\sqrt{\frac{x}{1+x^{2}}}.\]
We have
\begin{align*}
\frac{d}{dx}\left [\left (\frac{x}{1+x^{2}}\right )^{\frac{1}{2}}\right ] & =\frac{1}{2}\left (\frac{x}{1+x^{2}}\right )^{-\frac{1}{2}}\frac{d}{dx}\left (\frac{x}{1+x^{2}}\right )\\
& =\frac{1}{2}\left (\frac{x}{1+x^{2}}\right )^{-\frac{1}{2}}\frac{(1+x^{2})(1)-x(2x)}{(1+x^{2})^{2}}\\
& =\frac{1-x^{2}}{2(1+x^{2})\sqrt{x(1+x^{2})}}.
\end{align*}
The result holds for all \(x>0\). \(\sharp\)
\begin{equation}{\label{d}}\tag{D}\mbox{}\end{equation}
The Mean-Value Theorem.
Theorem. (Rolle’s Theorem). Let \(f\) be differentiable on the open interval \((a,b)\) and continuous on the closed interval \([a,b]\). Suppose that \(f(a)=f(b)=0\). Then, there is at least one number \(c\in (a,b)\) satisfying \(f'(c)=0\).
Theorem. (Mean-Value Theorem). Suppose that \(f\) is differentiable on the open interval \((a,b)\) and continuous on the closed interval \([a,b]\). Then, there is at least one number \(c\in (a,b)\) satisfying
\[f'(c)=\frac{f(b)-f(a)}{b-a}\]
or, equivalently,
\[f(b)-f(a)=f'(c)(b-a).\]
We can see that Rolle’s theorem is a special case of the mean-value theorem. The Rolle’s theorem is sometimes stated as:
Theorem. (Rolle’s Theorem). Let \(f\) be differentiable on the open interval \((a,b)\) and continuous on the closed interval \([a,b]\). If \(f(a)=f(b)\), then there is at least one number \(c\in (a,b)\) satisfying \(f'(c)=0\). \(\sharp\)
Increasing and Decreasing Functions.
A function \(f\) is said to increase on the interval \(I\) when, for every two numbers \(x_{1},x_{2}\in I\),
\[x_{1}<x_{2}\mbox{ implies }f(x_{1})<f(x_{2}),\]
and \(f\) is said to decrease on the interval \(I\) when, for every two numbers \(x_{1},x_{2}\in I\),
\[x_{1}<x_{2}\mbox{ implies }f(x_{1})>f(x_{2}).\]
Theorem. Let \(f\) be differentiable on an open interval \(I\). We have the following properties.
(i) If \(f'(x)>0\) for all \(x\in I\), then \(f\) increases on \(I\).
(ii) If \(f'(x)<0\) for all \(x\in I\), then \(f\) decreases on \(I\).
(iii) If \(f'(x)=0\) for all \(x\in I\), then \(f\) is constant on \(I\). \(\sharp\)
Suppose that \(f\) is differentiable on an open interval \(I\). Then \(f(x)=\alpha\) ($\alpha$ a constant) for all \(x\in I\) if and only if \(f'(x)=0\) for all \(x\in I\).
Theorem. Let \(f\) be continuous on an arbitrary interval \(I\) and differentiable on the interior of \(I\). We have the following properties.
(i) If \(f'(x)>0\) for all \(x\) in the interior of \(I\), then \(f\) increases on all of \(I\).
(ii) If \(f'(x)<0\) for all \(x\) in the interior of \(I\), then \(f\) decreases on all of \(I\).
(iii) If \(f'(x)=0\) for all \(x\) in the interior of \(I\), then \(f\) is constant on all of \(I\). \(\sharp\)
Example. The function \(f(x)=\sqrt{1-x^{2}}\) has derivative
\[f'(x)=-\frac{x}{\sqrt{1-x^{2}}}.\]
Since \(f'(x)>0\) for all \(x\in (-1,0)\) and \(f\) is continuous on \([-1,0]\), it follows that \(f\) increases on \([-1,0]\). Since \(f'(x)<0\) for all \(x\in (0,1)\) and \(f\) is continuous on \([0,1]\), it follows that \(f\) decreases on \([0,1]\). \(\sharp\)
Example. The function \(f(x)=1/x\) is defined for all \(x\neq 0\). The derivative
\[f'(x)=-\frac{1}{x^{2}}\]
is negative for all \(x\neq 0\). Therefore, the function \(f\) decreases both on \((-\infty ,0)\) and \((0,+\infty )\). \(\sharp\)
Example. Let \(f(x)=x-2\sin x\) for \(0\leq x\leq 2\pi\). Find the intervals on which \(f\) increases and the intervals on which \(f\) decreases. The derivative of \(f\) is \(f'(x)=1-2\cos x\). Setting \(f'(x)=0\), we have \(\cos x=1/2\). The solutions are \(x=\pi /3\) and \(x=5\pi /3\). Therefore \(f’\) has constant sign on the intervals \((0,\pi /3)\), \((\pi /3,\pi /5)\) and \((\pi /5,2\pi )\). We conclude that \(f\) decreases on \([0,\pi /3]\), increases on \([\pi /3,5\pi /3]\) and decreases
on \([5\pi /3,2\pi ]\). \(\sharp\)
Theorem. We have the following properties.
(i) Let \(I\) be an open interval. If \(f'(x)=g'(x)\) for all \(x\in I\), then \(f\) and \(g\) differ by a constant on \(I\).
(ii) Let \(I\) be an arbitrary interval. If \(f'(x)=g'(x)\) for all \(x\) in the interior of \(I\) and \(f\) and \(g\) are continuous on \(I\), then \(f\) and \(g\) differ by a constant on \(I\). \(\sharp\)
Example. Given \(f'(x)=6x^{2}-7x-5\) for all real \(x\) and \(f(2)=1\), find \(f\). We can obtain \(f(x)=2x^{3}-\frac{7}{2}x^{2}-5x+C\). Since \(f(2)=1\), we get
$C=9$, which says \(f(x)=2x^{3}-\frac{7}{2}x^{2}-5x+9\). \(\sharp\)
Local Extreme Values.
A function \(f\) is said to have a local maximum at \(c\) when \(f(c)\geq f(x)\) for all \(x\) sufficiently close to \(c\). A function \(f\) is said to have a local minimum at \(c\) when \(f(c)\leq f(x)\) for all \(x\) sufficiently close to \(c\). The local maxima and local minima of \(f\) are called the local extreme values of \(f\).
Theorem. Suppose that \(f\) has a local maximum or minimum at \(c\). Then, either \(f'(c)=0\) or \(f'(c)\) does not exist. \(\sharp\)
Given a function \(f\). The numbers \(c\) in the domain of \(f\) for which either \(f'(c)=0\) or \(f'(c)\) does not exist are called critical numbers of \(f\).
Example. In the case of
\begin{align*} f(x) & =|x+1|+2\\ & =\left\{\begin{array}{ll}
-x+1, & x<-1\\
x+3, & x\geq -1,
\end{array}\right .\end{align*}
differentiation gives
\[f'(x)=\left\{\begin{array}{ll}
-1, & x<-1\\
1, & x>-1.
\end{array}\right .\]
This derivative is never \(0\). It fails to exist only at \(-1\). The number \(-1\) is the only critical number. The value \(f(-1)=2\) is a local minimum. \(\sharp\)
Example. The function
\[f(x)=\left\{\begin{array}{ll}
-2x+5, & x<2\\
-\frac{1}{2}x+2, & x\geq 2
\end{array}\right .\]
is everywhere decreasing. Although \(2\) is a critical number ($f'(2)$ does not exist), \(f(2)=1\) is not a local extreme value. \(\sharp\)
Theorem. (The First Derivative Test). Suppose that \(c\) is a critical number of \(f\), and that \(f\) is continuous at \(c\). Then, we have the following properties.
(i) If there is a positive number \(\delta\) satisfying \(f'(x)>0\) for all \(x\in (c-\delta ,c)\) and \(f'(x)<0\) for all \(x\in (c,c+\delta )\), then \(f(c)\) is a local maximum.
(ii) If there is a positive number \(\delta\) satisfying \(f'(x)<0\) for all \(x\in (c-\delta ,c)\) and \(f'(x)>0\) for all \(x\in (c,c+\delta )\), then \(f(c)\) is a local minimum.
(iii) If there is a positive number \(\delta\) satisfying \(f'(x)\) keeps constant sign on \((c-\delta ,c)\cup (c,c+\delta )\), then \(f(c)\) is not a local extreme value. \(\sharp\)
The requirement that \(f\) should be continuous at \(c\) in the hypothesis of the above theorem is necessary. For example, \(1\) is a critical number of the function
\[f(x)=\left\{\begin{array}{ll}
1+2x, & x\leq 1\\
5-x, & x>1
\end{array}\right .\]
since \(f'(1)\) does not exist. Note that \(f'(x)=2>0\) to the left of \(1\) and \(f'(x)=-1<0\) to the right of \(1\), but \(f(1)=3\) is not a local maximum of \(f\).
Theorem. (The Second Derivative Test). Suppose that \(f'(c)=0\), and that \(f”(c)\) exists. If \(f”(c)>0\), then \(f(c)\) is a local minimum value. If \(f”(c)<0\), then \(f(c)\) is a local maximum value. \(\sharp\)
Example. For \(f(x)=2x^{3}-3x^{2}-12x+5\), we have \(f'(x)=6(x-2)(x+1)\) and \(f”(x)=12x-6\). The critical numbers are \(2\) and \(-1\). Since \(f”(2)=18>0\)
and \(f”(-1)=-18<0\), we conclude that \(f(2)=-15\) is a local minimum and \(f(-1)=12\) is a local maximum. \(\sharp\)
For functions defined on an open interval or on the union of open intervals the critical numbers are those where the derivative is \(0\) or the derivative does not exist. For functions defined on a closed or half-closed interval \([a,b]\), \([a,b)\), \((a,b]\), \([a,+\infty )\) or \((-\infty ,b]\) or on a union of such intervals, the endpoints of the domain ($a$ and \(b\) in the case of \([a,b]\), \(a\) in the case of \([a,b)\) or \([a,+\infty )\) and so forth) are also called critical numbers. Endpoints can give rise to what are called endpoint maxima and endpoint minima. If \(c\) is an endpoint of the domain of \(f\), then \(f\) is said to have an endpoint maximum at \(c\) when \(f(c)\geq f(x)\) for all \(x\) in the domain of \(f\) sufficiently close to \(c\). It is said to have an endpoint minimum at \(c\) when \(f(c)\leq f(x)\) for all \(x\) in the domain of \(f\) sufficiently close to \(c\). A function \(f\) is said to have an absolute maximum at \(d\) when \(f(d)\geq f(x)\) for all \(x\) in the domain of \(f\). Also, \(f\) is said to have an absolute minimum at \(d\) when \(f(d)\leq f(x)\) for all \(x\) in the domain of \(f\). These values are called the absolute extreme values of \(f\). Here \(d\) can be an interior point or an endpoint.
Theorem. Suppose that \(f\) is continuous on a closed, bounded interval \([a,b]\). Then \(f\) is bounded on \([a,b]\), and \(f\) attains its (absolute) maximum value \(M\) and its (absolute) minimum value \(m\) on \([a,b]\). \(\sharp\)
The absolute extreme values of \(f\) on \([a,b]\) can be determined as the following steps
Step 1. Find the critical numbers \(c_{1},c_{2},\cdots ,c_{n}\) of \(f\) in the open interval \((a,b)\).
Step 2. Calculate \(f(a),f(c_{1}),f(c_{2}),\cdots ,f(c_{n}),f(b)\).
Step 3. The largest of the numbers found in Step 2 is the absolute maximum of \(f\) and the smallest is the absolute minimum of \(f\).
Example. Find the critical numbers and classify all extreme values of the function \(f\) defined on \([-\frac{1}{2},\frac{3}{2}]\) by
\begin{align*} f(x) & =x^{2}-2|x|+2\\ & =\left\{\begin{array}{ll}
x^{2}+2x+2, & -\frac{1}{2}\leq x<0\\
x^{2}-2x+2, & 0\leq x\leq\frac{3}{2}.
\end{array}\right .\end{align*}
Since \(f\) is continuous on a closed, bounded interval, we know that it has an absolute maximum and absolute minimum. This function is differentiable on the open interval \((-\frac{1}{2},\frac{3}{2})\), except at \(x=0\):
\[f'(x)=\left\{\begin{array}{ll}
2x+2, & -\frac{1}{2}<x<0\\
2x-2, & 0<x<\frac{3}{2}\\
\mbox{does not exists}, & x=0.
\end{array}\right .\]
This makes \(0\) a critical number. Since \(f'(x)=0\) at \(x=1\), \(1\) is a critical number. The endpoints \(-\frac{1}{2}\) and \(\frac{3}{2}\) are also critical numbers. (Note that \(2x+2=0\) when \(x=-1\), but we have ignored this value since it is not in the domain of \(f\).) Since \(f'(x)\) is positive on \((-\frac{1}{2},0)\) and \((1,\frac{3}{2})\), and negative on \((0,1)\), \(f\) increases on \((-\frac{1}{2},0)\) and \((1,\frac{3}{2})\), and decreases on \((0,1)\). Therefore, the results are summarized below.
\begin{align*}
& \mbox{$f(-\frac{1}{2})=\frac{5}{4}$ is an endpoint minimum};\\
& \mbox{$f(0)=2$ is a local maximum};\\
& \mbox{$f(1)=1$ is a local minimum};\\
& \mbox{$f(\frac{3}{2})=\frac{5}{4}$ is an endpoint maximum}.
\end{align*}
Also \(f(1)=1\) is the absolute minimum and \(f(0)=2\) is the absolute maximum. \(\sharp\)
A Summary on Finding All the Extreme Values of a Continuous Function \(f\):
(i) Find the critical numbers. These are the endpoints of the domain and the interior numbers \(c\) at which \(f'(c)=0\) or \(f'(c)\) does not exist.
(ii) Test each endpoint of the domain by examining the sign of the first derivative nearby.
(iii) Test each interior critical number \(c\) by examing the sign of the first derivative on both sides of \(c\) (first-derivative test) or by checking the sign of the second derivative at \(c\) itself (second-derivative test).
(iv) If \(f\) is defined on \((a,+\infty )\) for some number \(a\) or on \((-\infty ,b)\) for some number \(b\), determine the behavior of \(f\) as \(x\rightarrow +\infty\) or as \(x\rightarrow -\infty\).
(v) Determine whether any of the endpoint extremes and local extremes are absolute extremes.
Example. Find the critical numbers and classify all the extreme values of \(f(x)=\sin x-\sin^{2} x\) for \(x\in [0,2\pi ]\). The endpoints \(0\) and \(2\pi\) are critical numbers. On \((0,2\pi )\), we have
\begin{align*}
f'(x) & =\cos x-2\sin x\cos x\\
& =\cos x(1-2\sin x).
\end{align*}
Setting \(f'(x)=0\), the equation is satisfied when \(\cos x=0\), which gives \(x=\pi /2\) and \(x=3\pi /2\), and when \(\sin x=\frac{1}{2}\), which gives \(x=\pi /6\) and \(x=5\pi /6\). Therefore, the numbers \(\pi /6, \pi /2, 5\pi /6\) and \(3\pi /2\) are interior critical numbers. Since \(f’\) is positive on intervals \((0,\pi /6)\), \((\pi /2,5\pi /6)\), \((3\pi /2.2\pi )\) and decreases on \((\pi /6,\pi /2 )\) and \((5\pi /6,3\pi /2)\). Therefore, the results are summarized below.
\begin{align*}
& \mbox{$f(0)=0$ is an endpoint minimum};\\
& \mbox{$f(\pi /6)=\frac{1}{4}$ is a local maximum};\\
& \mbox{$f(\pi /2)=0$ is a local minimum};\\
& \mbox{$f(5\pi /6)=\frac{1}{4}$ is a local maximum};\\
& \mbox{$f(3\pi /2)=-2$ is a locaal minimum};\\
& \mbox{$f(2\pi )=0$ is an endpoint maximum}.
\end{align*}
Also, \(f(\pi /6)=f(5\pi /6)=\frac{1}{4}\) is the absolute maximum of \(f\) and \(f(3\pi /2)=-2\) is the absolute minimum. \(\sharp\)
