Question

pe graph of the function /is shown, and consists of two line segments. Let g be the functions given by s6o - Trend a) Find g(-1), g'(-1), "(-1). f fis Stadt رک 03 9 -2 -10 2 b) For what values of x in the open interval (-2, 2) is g increasing? Explain your reasoning from - tol on the axis bus. interval it se starts at an increase then. a declace to finchis the graph. -1 down ito coot -2 c) For what values of x in the open interval (-2, 2) is the graph of g concave down? Explain your reasoning. 012 (-2.-3) -37 (2.-3) Graph of 이 d) On the axes provided, sketch the graph of g on the closed interval (-2,2].

Answer 1

a)

We have the following function given by an integral

$g(x)=\int ^x_0f(t)dt$

Let's evaluate it at g(-1), we have

$g(-1)=\int ^{-1}_0f(t)dt$

Looking at the graph of f(t) we can see that it's the area of a triangle of base 1 and height 3, therefore

$g(-1)=\int ^{-1}_0f(t)dt=-\int ^0_{-1}f(t)dt=-\frac{3\cdot1}{2}=-\frac{3}{2}$

g(-1) is -3/2

$g(-1)=\int ^{-1}_0f(t)dt=-\frac{3}{2}$

To find the derivative of f(x) we can use the fundamental theorem, therefore

$g^{\prime}(x)=\frac{d}{dx}\int ^x_0f(t)dt=\frac{d}{dx}\lbrack F(x)-F(0)\rbrack=f(x)$

Therefore

$g^{\prime}(x)=f(x)$

Looking at the graph we can see that, g'(-1) = 0

$g^{\prime}(-1)=f(-1)=0$

And now for the last derivative, we have

$g^{\prime}^{\prime}(-1)=f^{\prime}(-1)$

As we can see f(x) is a line of positive slope at the interval (-∞, 0], the derivative of a line if the slope, we can easily see that the slope of f(x) at (-∞, 0] is 3, therefore

$g^{\prime\prime}(-1)=f^{\prime}(-1)=3$

d) Here we will use the fact that f is compounded function (or a modular function), so let's find the expression of f and then do the integral to find g, as we can see, we have a line for (-∞, 0] and another line for [0, ∞), the equations are very easy to find.

For (-∞, 0] we have

$y=3x+3$

And for [0, ∞)

$y=-3x+3$

Then we can write the function f

$f(x)=\begin{cases}3x+3,\quad x<0 \\ -3x+3,\quad x\ge0\end{cases}$

Therefore when we solve the integral for g(x) we will find two different functions, one for x>0 and the other for x<0. then let's do it first for x<0

$\begin{gathered} \int ^x_0f(t)dt,\quad x<0 \\ \\ \int ^x_0(3t+3)dt\Rightarrow\frac{3t^2}{2}+3t+c\Rightarrow\frac{3x^2}{2}+3x+c-\frac{3\cdot0^2}{2}-3\cdot0-c \\ \\ \\ \int ^x_0(3t+3)dt=\frac{3x^2}{2}+3x \end{gathered}$

Now we will repeat the exact same process but now for x > 0

$\begin{gathered} \int ^x_0f(t)dt,\quad x>0 \\ \\ \int ^x_0(-3t+3)dt\Rightarrow-\frac{3t^2}{2}+3t+c\Rightarrow-\frac{3x^2}{2}+3x+c+\frac{3\cdot0^2}{2}-3\cdot0-c \\ \\ \int ^x_0(-3t+3)dt=-\frac{3x^2}{2}+3x \end{gathered}$

Therefore we have the expression for g now

$\begin{gathered} g(x)=\begin{cases}\frac{3x^2}{2}+2x,\quad x<0 \\ \\ -\frac{3x^2}{2}+2x,\quad x\ge0\end{cases} \\ \end{gathered}$

Then to graph g we must graph two quadratics! Let's plot these quadratics from -2 to 2