Looking at the graph
we have that
For the interval (-3,0) the slope is negative ---> function f(x) is decreasing
For the interval (0,2) the slope of the first derivative is positive ---> the function f(x) is increasing
so
at x=0 there is a minimum value
The equation of the derivative at the interval (-3,0) is y'=-x
The equation of the derivative at the interval (0,2) is y'=x
The equation of the derivative at the interval (2,4) is y'=2x
The equation of the derivative at the interval (4,5) is
we have the points
(4,2) and (5,0)
Find out the slope
m=(0-2)/(5-4)
m=-2/1
m=-2
Find out the equation of the line
y=mx+b
m=-2
point (5,0)
substitute and solve for b
0=-2(5)+b
b=10
therefore
y'=-2x+10
Find out the function f(x) at the interval (4,5)
Remember that
f(5)=10
Find out the value of C
For x=5
-(5)^2+10(5)+C=10
-25+50+C=10
C=-15
therefore
f(x)=-x^2+10x-15 -------------> interval (4,5)
For x=4 ------> f(x)=-(4)^2+10(4)-15=-16+40-15=9
Part 2
Find out the equation of the function f(x) at the interval (2,4)
y'=2 -------> applying the integral ------> f(x)=2x+C
Remember that
For x=4 ----> f(x)=9
Find out the value of C
9=2(4)+C
C=1
therefore
f(x)=2x+1 -------------> interval (2,4)
For x=2 -----> f(x)=2(2)+1=5
Part 3
Find out the equation of f(x) at the interval (0,2)
y'=x ----> applying the integral -----> f(x)=(1/2)x^2+C
Remember that
For x=2 ---> f(x)=5
5=(1/2)(2)^2+C
5=2+C
C=3
therefore
f(x)=(1/2)x^2+3 ------> interval (0,2)
For x=0 ---------> f(x)=3
Part 4
Find out the equation of f(x) at the interval (-3,0)
y'=-x ------> f(x)=-(1/2)x^2+C
For x=0 ------> f(x)=3
so
3=-(1/2)(0)^2+C
C=3
therefore
f(x)=-(1/2)x^2+3 ------> interval (-3,0)
Part 5
Graph the given functions
The absolute minimum value is f(x)=3 at x=0