Chapter 2.7, Problem 25E

To determine

To calculate: The first and second derivative of the function $f\left(x\right)=10x^2-50x$ for $-5\le x\le 5$ also sketch the graph of the function.

Answer to Problem 25E

The value of first derivative of the function is $f\text{'}\left(x\right)=20x-50$ and second derivative is $f\text{'}\text{'}\left(x\right)=20$ . The graph of the function $f\left(x\right)=10x^2-50x$ for $-5\le x\le 5$ is,

  

Explanation of Solution

Given information:

The function $f\left(x\right)=10x^2-50x$ .

Formula used:

Let a function g be continuous on closed interval $\left[c,d\right]$ and differentiable on open interval $\left(c,d\right)$ ,

If first derivative of the function is greater than zero that is $g\text{'}\left(x\right)>0$ for every x in $\left(c,d\right)$ , then the function g is increasing on interval $\left[c,d\right]$ .

If first derivative of the function is less than zero that is $g\text{'}\left(x\right)<0$ for every x in $\left(c,d\right)$ , then the function g is decreasing on interval $\left[c,d\right]$ .

If second derivative of the function is greater than zero that is $g\text{'}\text{'}\left(x\right)>0$ for every x in $\left(c,d\right)$ , then the function g is concave up on interval $\left[c,d\right]$ .

If second derivative of the function is less than zero that is $g\text{'}\text{'}\left(x\right)<0$ for every x in $\left(c,d\right)$ , then the function g is concave down on interval $\left[c,d\right]$ .

The point where the graph changes it nature is known as the point of inflection.

Power rule of differentiation, $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ ,

Calculation:

Consider the provided function $f\left(x\right)=10x^2-50x$ .

Evaluate the first derivative of the function,.

Apply sum rule of differentiation,

  $\begin{array}{c}\frac{d}{dx}f\left(x\right)=\frac{d}{dx}\left(10x^2-50x\right)\\ =\frac{d}{dx}\left(10x^2\right)+\frac{d}{dx}\left(-50x\right)\end{array}$

Apply the power rule of differentiation, $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ .

  $\begin{array}{c}\frac{d}{dx}f\left(x\right)=\frac{d}{dx}\left(10x^2-50x\right)\\ =\frac{d}{dx}\left(10x^2\right)+\frac{d}{dx}\left(-50x\right)\\ =10\cdot 2x^{2-1}-50\\ =20x-50\end{array}$

Evaluate the second derivative of the function, differentiate the first derivative again with respect to x .

Apply the power rule of differentiation, $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ .

  $\begin{array}{c}\frac{d^2}{d{x}^2}f\left(x\right)=\frac{d}{dx}\left(20x-50\right)\\ =\frac{d}{dx}\left(20x\right)-\frac{d}{dx}\left(50\right)\\ =20-0\\ =20\end{array}$

Recall if first derivative of the function is greater than zero that is $g\text{'}\left(x\right)>0$ for every x in $\left(c,d\right)$ , then the function g is increasing on interval $\left[c,d\right]$ .

If first derivative of the function is less than zero that is $g\text{'}\left(x\right)<0$ for every x in $\left(c,d\right)$ , then the function g is decreasing on interval $\left[c,d\right]$ .

If second derivative of the function is greater than zero that is $g\text{'}\text{'}\left(x\right)>0$ for every x in $\left(c,d\right)$ , then the function g is concave up on interval $\left[c,d\right]$ .

If second derivative of the function is less than zero that is $g\text{'}\text{'}\left(x\right)<0$ for every x in $\left(c,d\right)$ , then the function g is concave down on interval $\left[c,d\right]$ .

The point where the graph changes it nature is known as the point of inflection.

To sketch the graph of the function $f\left(x\right)=10x^2-50x$ follow the steps below,

Observe that first derivative of the function $f\text{'}\left(x\right)=20x-50$ .

Now, first derivative can be both negative and positive depending on values of x ,

First derivative will be negative if,

  $\begin{array}{c}20x-50<0\\ 20x<50\\ x<\frac{50}{20}\\ x<2.5\end{array}$

Therefore, the function $f\left(x\right)=10x^2-50x$ is decreasing when $x<2.5$ .

First derivative will be positive if,

  $\begin{array}{c}20x-50\ge 0\\ 20x\ge 50\\ x\ge \frac{50}{20}\\ x\ge 2.5\end{array}$ ~

Therefore, the function $f\left(x\right)=10x^2-50x$ is increasing when $x\ge 2.5$ .

Next observe that second derivative of the function $f\text{'}\text{'}\left(x\right)=20$ is positive for $-5\le x\le 5$ , so $f\left(x\right)=10x^2-50x$ is concave up.

At $x=2.5$ , the function changes it nature so it is a point of inflection.

Therefore, the graph of the function $f\left(x\right)=10x^2-50x$ for $-5\le x\le 5$ is provided below,

  

Thus, the value of first derivative of the function is $f\text{'}\left(x\right)=20x-50$ and second derivative is $f\text{'}\text{'}\left(x\right)=20$ .