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(dU)/(dx)+U(dV)/(dx)=0`

Where:

`\begin{gathered} U=x \\ V=\cos (x^2+1) \\ (dU)/(dx)=1 \\ \\ (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).