Find the exact coordinates of the inflection points and critical points of the function f(x) =...

Question:

Find the exact coordinates of the inflection points and critical points of the function

{eq}f(x) = 4x^5 - 100x^3 - 5 {/eq}

Critical point of a Function:

First we have to understand the critical point to solve this problem:

The point where {eq}f'(x) = 0 {/eq} or {eq}f'(x) {/eq} is undefined but {eq}f(x) {/eq} is defined that points are called the critical points of {eq}f(x) {/eq}.

Note: The point where {eq}f''(x) = 0 {/eq}, that point is called the inflection point of the function.

Answer and Explanation:


Given:

{eq}\displaystyle f(x) = 4 x^5 - 100 x^3 - 5 {/eq}

We will compute the derivative of {eq}f(x) {/eq} to get the critical point.

Take the derivative of {eq}f(x) {/eq} with respect to {eq}x {/eq}:

$$\begin{align*} \displaystyle f'(x) &= \frac{d}{dx} \left [ 4 x^5 - 100 x^3 - 5 \right ] \\ &= 4 \times 5 x^4 - 100 \times 3 x^2 - 0 &\text{(Differentiating using the formula } \frac{d}{dx}[x^n] = n x^{n - 1} \text{)}\\ &= 20 x^4 - 300 x^2 \\ f'(x) &= 20 x^4 - 300 x^2 \ \ ..... (1) \end{align*} $$

Plug in {eq}f'(x) = 0 {/eq} for getting the critical point:

$$\begin{align*} \displaystyle 20 x^4 - 300 x^2 &= 0\\ 20 x^2 (x^2 - 15) &= 0 \\ 20 x^2 &= 0 \text{ or } x^2 - 15 = 0 \\ x &= 0 \text{ or } x^2 = 15 \\ x &= 0 \text{ or } x = \pm \sqrt{15} \end{align*} $$


Thus, we have three critical points of {eq}f(x) {/eq} at {eq}x = 0, \sqrt{15}, - \sqrt{15} {/eq}

As we know the point where {eq}f''(x) = 0 {/eq}, that point is called the inflection point.

So first we will compute {eq}f''(x) {/eq}.

Take the derivative of {eq}f'(x) {/eq} with respect to x:

$$\begin{align*} \displaystyle f''(x) &= \frac{d}{dx}[20 x^4 - 300 x^2 ] \\ &= 20 \times 4 x^3 - 300 \times 2 x &\text{(Differentiating using the formula } \frac{d}{dx}[x^n] = n x^{n - 1} \text{)}\\ f''(x) &= 80 x^3 - 600 x \\ f''(x) &= 0 &\text{(Plugging in } f''(x) = 0 \text{ for getting the inflection point )}\\ 80 x^3 - 600 x &= 0 \\ 40 x (2 x^2 - 15) &= 0 \\ 40 x &= 0 \text{ or } 2 x^2 - 15 = 0 \\ x &= 0 \text{ or } 2 x^2 = 15 \\ x &= 0 \text{ or } x^2 = \frac{15}{2} \\ x &= 0 \text{ or } x = \pm \sqrt{\frac{15}{2}} \end{align*} $$

Thus the inflection points are: {eq}\displaystyle x = 0, \sqrt{\frac{15}{2}}, - \sqrt{\frac{15}{2}} {/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
198K

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