Chapter 3.4, Problem 10QR

To determine

To calculate: The value of $\frac{d^{2}y}{d{x}^2}\text{ at }x=7$ by analysing the equation algebraically.

Answer to Problem 10QR

The value of $\frac{d^{2}y}{d{x}^2}\text{ at }x=7$ is -32.

Explanation of Solution

Given Information: The equation of the quadratic function is $f\left(x\right)=-16x^2+160x-256$

Calculation:

Consider the equation $f\left(x\right)=-16x^2+160x-256$.

The derivative of the function is,

  $\begin{array}{c}y=-16x^2+160x-256\\ \frac{dy}{dx}=\frac{d}{dx}\left(-16x^2+160x-256\right)\\ =-32x+160\end{array}$

The second derivative of $f\left(x\right)=-16x^2+160x-256$ ,

  $\begin{array}{c}\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)\\ =\frac{d}{dx}\left(-32x+160\right)\\ =-32\end{array}$

This implies that for all values of x, the value of $\frac{d^{2}y}{d{x}^2}=-32$.

Therefore,

  $\frac{d^{2}y}{d{x}^2}{|}_{x=7}=-32$

Conclusion:

The value of $\frac{d^{2}y}{d{x}^2}\text{ at }x=7$ is -32.