Question

The derivative of the function is given by ‘(X) = Xcos (X squared plus one) How many points of the inflection does the graph F have on the open interval negative two less than X less than two

Answer 1

The Solution:

given the derivative function below:

$f^{\prime}(x)=x\cos (x^2+1)$

We are required to find the number of points of inflection of the graph of the function f(x) on the open interval -2 < x < 2

Recall:

At the point of inflection of a function, the second derivative of the function is equal to zero. That is,

$f^{\doubleprime}(x)=0$

Differentiating f'(x) with respect to x using the Product Rule, we get

$f^{\prime}\text{'(x)}=V\frac{dU}{dx}+U\frac{dV}{dx}=0$

Where:

$\begin{gathered} U=x \\ V=\cos (x^2+1) \\ \frac{dU}{dx}=1 \\ \\ \frac{dV}{dx}=-2x\sin (x^2+1) \end{gathered}$

Substituting, we get

$f^{\prime}\text{'(x)}=1\cdot\cos \mleft(x^2+1\mright)+\mleft(-\sin \mleft(x^2+1\mright)\cdot\: 2x\mright)x=0$$f^{\prime}\text{'(x)}=\cos \mleft(x^2+1\mright)-2x^2\sin \mleft(x^2+1\mright)=0$

To find the number of points of inflection of the function, we shall solve graphically for x .

$\cos (x^2+1)=2x^2\sin (x^2+1)$

Solving graphically using the Desmos App, we have

From the graph, the points of inflection in the open interval - 2 < x < 2 exist at the points below:

$(-1.5333,-0.978)\text{ and (1.5333,-0.978)}$

Thus, the number of points of inflection of the function in the given interval is 2.

Therefore, the correct answer is two (2).