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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) = \frac{2}{3}x^3 -2x^2 - 6x {/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}.

Inflection point: 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) = \frac{2}{3} x^3 - 2 x^2 - 6 x {/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}:

$$\begin{align*} \displaystyle f'(x) &= \frac{d}{dx} \left [ \frac{2}{3} x^3 - 2 x^2 - 6 x \right ] \\ &= \frac{2}{3} \times 3 x^2 - 2 \times 2 x - 6 \times 1 &\text{(Differentiating using the formula } \frac{d}{dx}[x^n] = n x^{n - 1} \text{)}\\ f'(x) &= 2 x^2 - 4 x - 6 \ \ ..... (1) \end{align*} $$

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

$$\begin{align*} \displaystyle 2 x^2 - 4 x - 6 &= 0\\ x^2 - 2 x - 3 &= 0 &\text{(Dividing both sides by 2)}\\ x^2 - 3 x + x - 3 &= 0 \\ x (x - 3) + 1 (x - 3) &= 0 \\ (x - 3) (x + 1) &= 0 \\ x - 3 &= 0 \text{ or } x + 1 = 0 \\ x &= 3 \text{ or } x = - 1 \\ \end{align*} $$


Thus, we have two critical points of {eq}f(x) {/eq} at {eq}x = 3. - 1 {/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:

$$\begin{align*} \displaystyle f''(x) &= \frac{d}{dx}[2 x^2 - 4 x - 6 ] \\ &= 2 \times 2 x - 4 \times 1 - 0 &\text{(Differentiating using the formula } \frac{d}{dx}[x^n] = n x^{n - 1} \text{)}\\ f''(x) &= 4 x - 4 \\ f''(x) &= 0 &\text{(Plugging in } f''(x) = 0 \text{ for getting the inflection point )}\\ 4 x - 4 &= 0 \\ 4 x &= 4 \\ x &= 1 &\text{(Dividing both sides by 4)} \end{align*} $$

Thus the inflection points is: {eq}x = 1 {/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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