Q. 1C - Calculus

Question

Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample. 

(a) True or False: : If $f\text{'}\left(x\right)<0 for all x\in (0,3)$, then $f$ is decreasing on $[0,3]$.

(b) True or False: : If $f$ is increasing on $(-2,2)$, then $f\text{'}\left(x\right)\ge 0$ for all $x\in (-2,2)$.

(c) True or False: If $f\text{'}\left(x\right)=2x, then f\left(x\right)=x^2+C$for some constant $C$ 

(d) True or False: : If $f\text{'}$ is continuous on $(1,8)$ and $f\text{'}\left(3\right)$ is negative, then $f\text{'}$is negative on all of $(1,8)$.

(e) True or False: : If $f\text{'}$ changes sign at $x=3$, then $f\text{'}\left(3\right)=0$ 

(f) True or False: : If $f\text{'}(-2)=0, then f$ has either a local maximum or a local minimum at $x=-2$ 

(g) True or False: If $x=1$ is the only critical point of $f$ and $f\text{'}\left(0\right)$ is positive, then $f\text{'}\left(2\right)$ must be negative.

(h) True or False: If $f\text{'}\left(1\right)$ is negative and $f\text{'}\left(3\right)$ is positive, then $f$ has a local minimum at $x=2$.

Step-by-Step Solution

Answer

Part (a). True.

Part (b). True.

Part (c). True.  

Part (d). False.

Part (e). True.

Part (f). True.

Part (g). False.

Part (h). False.

Part (a) Step 1. Explanation

The slope of a function $f$ can be defined by it’s first derivative $f\text{'}\left(x\right)$.

In the given interval for $[0,3]$ the function is decreasing when slope of the graph is $f\text{'}\left(x\right)<0.$

Thus, the statement is true.

Part (b) Step 1. Explanation

The slope of a function $f$ can be defined by it’s first derivative $f\text{'}\left(x\right)$.

In the given interval $(-2,2)$ for the function is increasing when slope of the graph is $f\text{'}\left(x\right)\ge 0$.

Thus, the statement is true.

Part (c) Step 1. Explanation

If $f\text{'}\left(x\right)=2x,$ then by integrating $f\text{'}.$

$$\begin{gathered} f\text{'}\left(x\right)=2x \\ \int f\text{'}\left(x\right)dx=\int 2xdx \\ f\left(x\right)=2\left[\frac{x^2}{2}\right]+C \\ f\left(x\right)=x^2+C \end{gathered}$$

for some constant.

Thus, the statement is true.

Part (d) Step 1. Explanation

Consider the counter example $f\left(x\right)=10+4x^2-x^3$.

Then the first derivative is $f\text{'}\left(x\right)=8x-3x^2$.

In the given interval on,  $2,3\in (1,8)$

$$\begin{gathered} f\left(x\right)=10+4x^2-x^3 \\ f\text{'}\left(x\right)=8x-3x^2 \\ f\text{'}\left(2\right)=16-12 \\ f\text{'}\left(2\right)=4 \\ f\text{'}\left(2\right)>0 \end{gathered}$$

$$\begin{gathered} f\left(x\right)=10+4x^2-x^3 \\ f\text{'}\left(x\right)=8x-3x^2 \\ f\text{'}\left(3\right)=24-27 \\ f\text{'}\left(3\right)=-3 \\ f\text{'}\left(3\right)<0 \end{gathered}$$

So, $f\text{'}\left(3\right)$ is negative, then $f\text{'}$ is not negative on all  of $(1,8).$

Thus, the statement is false.

Part (e) Step 1. Explanation

Consider the example $f\left(x\right)=\frac{1}{27}{x}^3-x$

The first derivative is,

$$\begin{gathered} f\text{'}\left(x\right)=\frac{x^2}{9}-1 \\ \frac{x^2}{9}-1=0 \\ (x+3)(x-3)=0 \end{gathered}$$

When $f\text{'}$ changes sign at $x=3$, there exists a critical point at $x=3.$

Thus, $f\text{'}\left(3\right)=0$

Thus, the statement is true.

Part (f) Step 1. Explanation

For any given function $f$, the first derivative $f\text{'}\left(x\right)$ gives the local minimum or local maximum at $f\text{'}\left(x\right)=0.$

So, if $f\text{'}(-2)=0$ has either a local maximum or local minimum at $x=-2$.

Thus, the statement is true.

Part (g) Step 1. Explanation

Consider the counter example $f\left(x\right)=x^2-2x$

The first derivative $f\text{'}\left(x\right)=2x-2$ can be used to find the critical point when $f\text{'}\left(x\right)=0=2x-2$.

Here, $x=1$is the only critical point of $f$ and $f\text{'}\left(0\right)$ is negative, then $f\text{'}\left(2\right)$ is positive.

Thus, the statement is false.

Part (h) Step 1. Explanation

Consider the counter example $f\left(x\right)=-x^2+4x$

The first derivative $f\text{'}\left(x\right)=-2x+4$ can be used to find the critical point when $f\text{'}\left(x\right)=0=-2x+4$

Here, $x=2$ is local minimum where $f\text{'}\left(1\right)$ is positive and $f\text{'}\left(3\right)$ is negative.

Thus, the statement is false.