# Find the points of inflection. f(x) = x^{4} - 24x^{2} A) (0,0) B) (-2\sqrt{3}, 144), (2\sqrt{3},...

## Question:

Find the points of inflection.

{eq}\displaystyle\; f(x) = x^{4} - 24x^{2} {/eq}

A) {eq}\displaystyle\;\left(0,0\right) {/eq}

B) {eq}\displaystyle\;\left(-2\sqrt{3}, 144\right), \;\left(2\sqrt{3}, 144\right) {/eq}

C) {eq}\displaystyle\;\left(-2,-80\right), \;\left(2,-80\right) {/eq}

D) {eq}\displaystyle\;\left(0,0\right), \;\left(-2,-80\right), \;\left(2,-80\right) {/eq}

## Point of Inflection of a Function

{eq}{/eq}

It is a point in the graph of a function at which the curvature changes from convex to concave or from concave to convex. Point of inflection can be determined by setting the second derivative of a function to zero and solving for the independent variable.

For example, consider a function y = f(x). To find its point(s) of inflection, we need to solve for:

$$\displaystyle f''(x) = 0 \\$$

Note: If a solution of the above equation makes the first derivative of f(x) equal to zero, it is not considered as a point of inflection.

{eq}{/eq}

We are given that:

$$\displaystyle f(x) = x^{4} - 24x^{2} \\$$

{eq}\displaystyle{\therefore f'(x) = \frac{d}{dx}( x^{4} - 24x^{2}) \\ \Rightarrow f'(x) = 4x^3 - 2 \times (24x) \\ \Rightarrow f'(x) = 4x^3 - 48 x \\ } {/eq}

To find the second derivative of f(x) :

{eq}\displaystyle{f''(x) = \frac{d}{dx}(f'(x)) \\ \Rightarrow f''(x) = \frac{d}{dx}(4x^3 - 48x) \\ \Rightarrow f''(x) = 3 \times 4x^2 - 48 \\ \Rightarrow f''(x) = 12x^2 - 48 \\ } {/eq}

To find the point(s) of inflection:

{eq}\displaystyle{f''(x) = 0 \\ \Rightarrow 12x^2 - 48 = 0 \\ \Rightarrow x^2 = 4 \\ \Rightarrow x = 2 \ , \ -2 \\ } {/eq}

Now, we need to check the values of f'(x) at these two points :

{eq}\displaystyle{f'(2) = 4 \times 2^3 - 48 \times 2 \\ \Rightarrow f'(2) = -64 \\ } {/eq}

{eq}\displaystyle{f'(-2) = 4 \times (-2)^3 - 48 \times (-2) \\ \Rightarrow f'(-2) = 64 \\ } {/eq}

We find that the first derivative is non-zero at both the points. Therefore, x = - 2 and x = 2 are the points of inflection of f(x).

To find the values of f(x) at x = - 2 and x = 2

{eq}\displaystyle{f(2) = 2^4 - 24 \times 2^2 \\ \Rightarrow f(2) = -80 \\ Also, \ f(2) = (-2)^4 - 24 \times (- 2)^2 \\ \Rightarrow f(2) = -80 \\ } {/eq}

Option C is correct. 