Chapter 4, Problem 1RE

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

1. a. If f′(c) = 0, then f has a local maximum or minimum at c.
2. b. If f″(c) = 0, then f has an inflection point at c.
3. c. F(x) = x² + 10 and G(x) = x² − 100 are antiderivatives of the same function.
4. d. Between two local minima of a function continuous on (−∞, ∞), there must be a local maximum.
5. e. The Linear approximation to f(x) = sin x at x = 0 is L(x) = x.
6. f. If $\underset{x\to \infty }{\lim }f(x)=\infty$ and $\underset{x\to \infty }{\lim }g(x)=\infty$, then $\underset{x\to \infty }{\lim }(f(x)-g(x))=0$.

To determine

Whether the statement, “If $f^{\prime }\left(c\right)=0$, then f has a local maximum or minimum at c” is true or false.

Answer to Problem 1RE

The statement is false.

Explanation of Solution

Definition used:

A function f has a local maximum at a point $x_0$ if the values $f\left(x\right)$ of f for x near $x_0$ are all less than $f\left(x_0\right)$.

A function f has a local minimum at a point $x_0$ if the values $f\left(x\right)$ of f for x near $x_0$ are all greater than $f\left(x_0\right)$.

Calculation:

The following example will disprove the given statement.

Consider the function $f\left(x\right)=x^3$.

Differentiate with respect to x.

$\begin{array}{l}\frac{d}{dx}f\left(x\right)=\frac{d}{dx}\left(x^3\right)\\ f^{\prime }\left(x\right)=3x^2\end{array}$

Substitute $x=0$.

$\begin{array}{c}{f}^{\prime }\left(0\right)=3\times 0^2\\ =0\end{array}$

Hence, $x=0$ is neither a local maximum nor a local minimum.

The point at $x=0$ is a critical point of f but is not necessarily local maximum or a local minimum.

Thus, the statement is false.

To determine

Whether the statement, “If $f^{\prime }\left(c\right)=0$ then f has an inflection point at c.” is true or false.

Answer to Problem 1RE

The statement is false.

Explanation of Solution

Definition used:

A point of inflection is a point on the curve at which the curvature changes its sign from positive to negative and vice versa.

Calculation:

Consider the function $f\left(x\right)=x^4$.

Differentiate f with respect to x.

$\begin{array}{l}\frac{d}{dx}f\left(x\right)=\frac{d}{dx}\left(x^4\right)\\ f^{\prime }\left(x\right)=4x^3\end{array}$

Differentiate $f^{\prime }\left(x\right)=4x^3$ with respect to x.

$\begin{array}{c}\frac{d}{dx}\left(f^{\prime }\left(x\right)\right)=\frac{d}{dx}\left(4x^3\right)\\ f^{″}\left(x\right)=4\times 3x^2\\ =12x^2\end{array}$

Substitute $x=0$ in $f^{\prime }\left(x\right)=4x^3$.

$\begin{array}{c}{f}^{\prime }\left(0\right)=4\times 0^3\\ =0\end{array}$

Since $f^{\prime }\left(0\right)=0$, the curvature does not changes its sign from positive to negative.

Hence, $x=0$ is not an inflection point.

Thus, the statement is false.

To determine

Whether the statement is, “$F\left(x\right)=x^2+10$ and $G\left(x\right)=x^2-100$ are anti-derivatives of the same function.” is true or false.

Answer to Problem 1RE

The statement is true.

Explanation of Solution

It is known that the antiderivative of a function f is a function whose derivative is f.

Note that a single continuous function can have infinite anti derivatives.

Also a family of antiderivatives, each of them, differs by a constant.

Thus, if F is an antiderivative of f then $G=F+c$ is also an antiderivative of f.

Note that F and G are in the same family of antiderivatives.

Therefore, $F\left(x\right)=x^2+10$ and $G\left(x\right)=x^2-100$ are antiderivatives of the same function, since it varies only by a constant.

Thus, the statement is true.

To determine

Whether the statement, “Between two local minima of a function continuous on $\left(-\infty ,\infty \right)$, there must be a local maximum.” is true or false.

Answer to Problem 1RE

The statement is true.

Explanation of Solution

Suppose that f is a continuous function that has local maxima at the points $x_1,x_2$.

Assume that there does not exist a local minimum in $\left(x_1,x_2\right)$.

If $x_1<x_2$, then $\inf \left\{f\left(x\right)/x\in \left[x_1,x_2\right]\right\}=f\left(x_1\right)\text{or}f\left(x_2\right)$.

Assume that $\inf \left\{f\left(x\right)/x\in \left[x_1,x_2\right]\right\}=f\left(x_1\right)$, then $f\left(x\right)>f\left(x_1\right)\text{when}x\in \left[x_1,x_2\right]$.

Also there exist $\delta >0$ such that $f\left(x_1\right)\in \left[x_1,x_1+\delta \right]$.

Then, $f\left(x\right)=f\left(x_1\right)$.

This is a contradiction.

Also, the function has a maximum on the closed interval determined by the two local minima, and the only way the maximum can occur at the endpoints is, if the function is constant, in which case every point is local max and min.

Since the function is continuous on $\left(-\infty ,\infty \right)$, there is at least one local maximum between two local minima.

Thus, the statement is true.

To determine

Whether the statement, “The linear approximation to $f\left(x\right)=\sin x$ at $x=0$ is $L\left(x\right)=x$.” is true or false.

Answer to Problem 1RE

The statement is true.

Explanation of Solution

Definition used:

Linear approximation

“Suppose f is differentiable on the interval I containing the point a. The linear approximation to f at a is a linear function $L\left(x\right)=f\left(a\right)+f^{\prime }\left(a\right)\left(x-a\right)$ for x in I.”

Calculation:

The function is $f\left(x\right)=\sin x$ and the point is $a=0$.

Calculate the derivative of $f\left(x\right)$ with respect to x.

$\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d\left(f\left(x\right)\right)}{dx}\\ =\frac{d\left(\sin x\right)}{dx}\\ =\cos x\end{array}$

Substitute $x=0$ in $f\left(x\right)=\sin x$.

$\begin{array}{c}f\left(0\right)=\sin 0\\ =0\end{array}$

Substitute $x=0$ in $f^{\prime }\left(x\right)=\cos x$.

$\begin{array}{c}{f}^{\prime }\left(0\right)=\cos 0\\ =1\end{array}$

Calculate the value of $f\left(x\right)$ at the point $a=0$.

Estimate the linear approximation to f near $a=0$ through definition mentioned above with $f\left(x\right)=\sin x$ and $f^{\prime }\left(x\right)=\cos x$.

$\begin{array}{c}L\left(x\right)=f\left(0\right)+f^{\prime }\left(0\right)\left(x-0\right)\\ =\sin 0+\cos \left(0\times x\right)\\ =0+\left(1\times x\right)\\ =x\end{array}$

Therefore, the equation of the line is $L\left(x\right)=x$.

Thus, the statement is true.

To determine

Whether the statement, “If $\underset{x\to \infty }{\lim }f\left(x\right)=\infty$ and $\underset{x\to \infty }{\lim }g\left(x\right)=\infty$, then $\underset{x\to \infty }{\lim }\left\{f\left(x\right)-g\left(x\right)\right\}=0$.” is true or false.

Answer to Problem 1RE

The statement is false.

Explanation of Solution

The following example will disprove the given statement.

Consider the function, $f\left(x\right)=x^2$.

Take $\underset{x\to \infty }{\lim }$ on both sides.

$\begin{array}{c}\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }{x}^2\\ =\infty \end{array}$

Consider the function, $g\left(x\right)=2x$.

Take $\underset{x\to \infty }{\lim }$ on both sides.

$\begin{array}{c}\underset{x\to \infty }{\lim }g\left(x\right)=\underset{x\to \infty }{\lim }2x\\ =\infty \end{array}$

Therefore, $f\left(x\right)-g\left(x\right)=x^2-2x$.

Take $\underset{x\to \infty }{\lim }$ on both sides,

$\begin{array}{c}\underset{x\to \infty }{\lim }\left\{f\left(x\right)-g\left(x\right)\right\}=\underset{x\to \infty }{\lim }\left\{x^2-2x\right\}\\ =\infty \end{array}$

Thus, the statement is false.