Answer:
Step-by-step explanation:
From the graph attached we will find the linear as well as exponential functions first.
Afterwards we will plug in the values of x to get the value of the function given.
For Linear function
It should be in the form of y = mx + c
We find c = 3
and two points passing through the line are (0, 3) and (2, 0)
so slope of the line should be [tex]m=\frac{y-y'}{x-x'}= \frac{3-0}{0-2}=-\frac{3}{2}[/tex]
Now we can say the linear function becomes [tex]f(x)=-\frac{3}{2}x+3[/tex]
For x = -3, [tex]f(3)=(-\frac{3}{2})(3)+3=(-\frac{9}{2})+3=(-\frac{3}{2})[/tex]
For x = -2 [tex]f(-2)=(-\frac{3}{2})(-2)+3=3+3=6[/tex]
For x = -1 [tex]f(-1)=(-\frac{3}{2})(-1)+3=\frac{3}{2}+3=\frac{9}{2}[/tex]
For x = 0 [tex]f(0)=3[/tex]
For x = 1 [tex]f(1)=-\frac{9}{2}[/tex]
For x = 2 f(2) = -6
For x = 3 [tex]f(3)=\frac{3}{2}[/tex]
Now for Exponential function
function will be in the form of [tex]y=a^{x}[/tex]
Since point (-1, 2) is passing through the exponential function
So [tex]2=(a)^{-1}=\frac{1}{a}[/tex]
⇒[tex]a=\frac{1}{2}[/tex]
Therefore exponential function is [tex]g(x)=(\frac{1}{2})^{x}[/tex]
Now from the given graph
g(-3) = 8
g(-2) = 6
g(-1) = 2
g(0) = 1
[tex]g(1)=\frac{1}{2}[/tex]
[tex]g(2)=(\frac{1}{2})^{2}=\frac{1}{4}=0.75[/tex]
[tex]g(3)=(\frac{1}{2})^{3}=\frac{1}{8}=0.125[/tex]
Now we will try to get the common values of x by analyzing the graphs of two functions.
we get the solutions for x as (-2.87) and (1.81)
