Chapter 6, Problem 60RE

(a)

To determine

To calculate:The value of g (-2) and g (4)

Answer to Problem 60RE

The value is $g\left(-2\right)=-4.5$ and $g\left(4\right)=2$

Explanation of Solution

Given information:

The graph of the function f consisting of three line segments, is shown below.

  

Formula used:

  $g\left(x\right)={\displaystyle \underset{1}{\overset{x}{\int }}f\left(t\right)}dt$

Calculation:

Since, it has been given that

  $g\left(x\right)={\displaystyle \underset{1}{\overset{x}{\int }}f\left(t\right)}dt$

Substitute the value of x equal to 4

  $\begin{array}{l}g\left(4\right)={\displaystyle \underset{1}{\overset{4}{\int }}f\left(t\right)}dt\\ =\left(\text{area}\text{of trapezium}\right)+\left(\begin{array}{l}\text{area of traingle}\\ \text{above the x-axis}\end{array}\right)-\left(\begin{array}{l}\text{area of traingle}\\ \text{below the x-axis}\end{array}\right)\\ =\left(\frac{1}{2}\times \left(1+3\right)\times 1\right)+\left(\frac{1}{2}\times \left(1\right)\times \left(1\right)\right)-\left(\frac{1}{2}\times \left(1\right)\times \left(1\right)\right)\\ g\left(4\right)=2\end{array}$

Substitute the value of x equal to -2 in equation of $g\left(x\right)$

  $\begin{array}{l}g\left(-2\right)={\displaystyle \underset{1}{\overset{-2}{\int }}f\left(t\right)}dt\\ =-{\displaystyle \underset{-2}{\overset{1}{\int }}f\left(t\right)}dt\\ =-\left(\text{area of traingle}\right)\\ =-\frac{1}{2}\times 3\times 3\\ g\left(-2\right)=-4.5\end{array}$

Conclusion:

The value is $g\left(-2\right)=-4.5$ and $g\left(4\right)=2$

(b)

To determine

To find:The instantaneous range of g with respect to x when x=2.

Answer to Problem 60RE

The value is $g\text{'}\left(2\right)=1$

Explanation of Solution

Formula used:

  $g\left(x\right)={\displaystyle \underset{1}{\overset{x}{\int }}f\left(t\right)}dt$

Calculation:

Since,

  $g\left(x\right)={\displaystyle \underset{1}{\overset{x}{\int }}f\left(t\right)}dt$

Take derivative both sides

  $\begin{array}{l}g\text{'}\left(x\right)=\frac{d}{dx}\left[{\displaystyle \underset{1}{\overset{x}{\int }}f\left(t\right)dt}\right]\\ =f\left(x\right)\end{array}$

Hence,

  $\begin{array}{l}g\text{'}\left(2\right)=f\left(2\right)\\ g\text{'}\left(2\right)=1\end{array}$

Conclusion:

The value is $g\text{'}\left(2\right)=1$

(c)

To determine

To find:The absolute minimum of g on the interval [-2,4]

Answer to Problem 60RE

The value is $x=2$ it is a point of minima.

Explanation of Solution

Formula used:

  $g\text{'}\left(x\right)=f\left(x\right)$

Calculation:

Since,

  $g\text{'}\left(x\right)=f\left(x\right)$

  $f\left(x\right)$ It is zero at two points at x equal to -2 and 3, so, these could be point of minima.

Now,

  $\begin{array}{l}g\left(3\right)={\displaystyle \underset{1}{\overset{3}{\int }}f\left(t\right)}dt\\ =\left(\text{area of trapexium}\right)+\left(\text{area of traingle}\right)\\ =\left(\frac{1}{2}\left(1+3\right)\left(1\right)\right)+\left(\frac{1}{2}\left(1\right)\left(1\right)\right)\\ g\left(3\right)=3\end{array}$

From part (a)

  $\begin{array}{l}g\left(-2\right)=-4.5\\ <g\left(3\right)\end{array}$

So, $x=2$ it is a point of minima.

Conclusion:

The value is $x=2$ it is a point of minima.

(d)

To determine

The value of g using x-ordinate.

Answer to Problem 60RE

At x = 1 double derivative does not change its sign, so, it is an inflexion point.

Explanation of Solution

Formula used:

The inflection point is used.

Calculation:

At inflection point, graph crosses the tangent, in other words double derivative change its sign.

At x = 2 double derivative does not change its sign, so, it is not an inflexion point.

At x = 1 double derivative does not change its sign, so, it is an inflexion point.

Conclusion:

At x = 1 double derivative does not change its sign, so, it is an inflexion point.