A differentiable function called f(x) achieves its maximum when x=0. Which of the following must...

Question:

A differentiable function called f(x) achieves its maximum when x=0. Which of the following must then be true?

1. The function {eq}\displaystyle p(x) = xf(x) {/eq} has a critical point when x = 0.

2. The function {eq}\displaystyle m(x) = (f(x))^2 {/eq} has its maximum at x = 0.

3. The function {eq}\displaystyle j(x) = f(x^2) {/eq} has its maximum at x = 0.

Maximum Points

The relative maximum points of a function {eq}\displaystyle y=f(x) {/eq} are the points {eq}\displaystyle c, {/eq} from the domain such that

{eq}\displaystyle f'(c)= 0 \text{ or } f'(x) -\text{ does not exist } {/eq} and the derivative changes from positive to negative at the point {eq}\displaystyle c. {/eq}

Answer and Explanation:

Knowing that {eq}\displaystyle f(x) {/eq} is differentiable (i.e. {eq}\displaystyle f'(x) {/eq} exists everywhere) and {eq}\displaystyle x=0 {/eq} is a local maximum, so {eq}\displaystyle f'(x) {/eq} changes from positive to negative at {eq}\displaystyle x=0. {/eq}

1. To determine the critical points of {eq}\displaystyle p(x)=xf(x) {/eq} we will obtain the first derivative.

{eq}\displaystyle \begin{align} p'(x)&=\frac{d}{dx}\left( xf(x)\right)\\ &=f(x)+xf'(x).\\ p'(0)&=f(0) -\text{ maximum value for }f(x) \implies \text{ does not guarantee that }p(x) =0 \text{ so the statement if FALSE} \end{align} {/eq}

2. To determine if {eq}\displaystyle x=0 {/eq} is a maximum point of {eq}\displaystyle m(x)=(f(x))^2 {/eq} we will calculate the first derivative and see if the derivative changes signs from positive to negative.

{eq}\displaystyle \begin{align} m'(x)&=\frac{d}{dx}\left( (f(x))^2\right)\\ &=2f(x)f'(x), &\left[\text{using Chain Rule}\right].\\ m'(0)&=2f(0) f'(0) =0 -\text{ because }f'(0)=0\\ \implies & m(x) \text{ has a critical point at } x=0, \text{ but it is not guaranteed that it is a maximum, because } \\ 2f(x)f'(x) & \text{ changes signs from } 2f(x) \text{ to } -2f(x), \text{which depends on th sign of }f(x),\\ &\boxed{\text{so, the statement is FALSE}}. \end{align} {/eq}

3. To determine if {eq}\displaystyle x=0 {/eq} is a maximum point of {eq}\displaystyle j(x)=f(x^2) {/eq} we will calculate the first derivative and see if the derivative changes signs from positive to negative.

{eq}\displaystyle \begin{align} j'(x)&=\frac{d}{dx}\left( f(x^2)\right)\\ &=2x f'(x^2), &\left[\text{using Chain Rule}\right].\\ j'(0)&=0 \\ \implies & j(x) \text{ has a critical point at } x=0\\ \\ \text{ and } 2xf'(x^2), & \text{ since }f'(x^2) <0 \text{ near zero, left or right at zero} \implies \text{ changes signs from positive ( at the left of zero, ) to negative (at the right of zero )}\\ &\text{ so, the point is local maximum }\\ &\boxed{\text{so, the statement is TRUE}}. \end{align} {/eq}


Learn more about this topic:

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Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9
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