Question

the figure above shows the graph of ( f ), the derivative of a twice - differentiable function ( f ), on the closed interval ( 0leq xleq8 ). the graph of ( f ) has horizontal tangent lines at ( x = 1 ), ( x = 3 ), and ( x = 5 ). the areas of the regions between the graph of ( f ) and the ( x ) - axis are labeled in the figure. the function ( f ) is defined for all real numbers and satisfies ( f(8)=4 ).
(a) find all values of ( x ) on the open interval ( 0lt xlt8 ) for which the function ( f ) has a local minimum. justify your answer.
(b) determine the absolute minimum value of ( f ) on the closed interval ( 0leq xleq8 ). justify your answer.
(c) on what open intervals contained in ( 0lt xlt8 ) is the graph of ( f ) both concave down and increasing? explain your reasoning.
(d) the function ( g ) is defined by ( g(x)=(f(x))^{3} ). if ( f(3)=-\frac{5}{2} ), find the slope of the line tangent to the graph of ( g ) at ( x = 3 ).

Explanation:

(a)

Step1: Recall the first - derivative test

A function \(y = f(x)\) has a local minimum at \(x = c\) if \(f^{\prime}(c)=0\) and \(f^{\prime}(x)\) changes sign from negative to positive at \(x = c\).
We know that \(f^{\prime}(x)\) has horizontal - tangent lines (i.e., \(f^{\prime}(x) = 0\)) at \(x = 1,x = 3,x = 5\).
For \(x = 1\):
To the left of \(x = 1\) (in an open interval around \(x = 1\)), \(f^{\prime}(x)<0\) (since the area between the curve \(y = f^{\prime}(x)\) and the \(x\) - axis is above the curve for \(x<1\) in the sense of the sign of \(f^{\prime}(x)\)). To the right of \(x = 1\) (in an open interval around \(x = 1\)), \(f^{\prime}(x)>0\) (because the function \(y = f^{\prime}(x)\) is above the \(x\) - axis for \(1For \(x = 3\):
To the left of \(x = 3\) (in an open interval around \(x = 3\)), \(f^{\prime}(x)>0\) (since \(y = f^{\prime}(x)\) is above the \(x\) - axis for \(1For \(x = 5\):
To the left of \(x = 5\) (in an open interval around \(x = 5\)), \(f^{\prime}(x)<0\) (since \(y = f^{\prime}(x)\) is below the \(x\) - axis for \(3

Step1: Use the fundamental theorem of calculus

The fundamental theorem of calculus states that \(f(x)-f(0)=\int_{0}^{x}f^{\prime}(t)dt\).
We know that \(f(8) = 4\). Also, \(f(x)-f(8)=\int_{8}^{x}f^{\prime}(t)dt\).
\(f(x)=f(8)+\int_{8}^{x}f^{\prime}(t)dt\).
\(\int_{0}^{1}f^{\prime}(t)dt=- 2\) (negative because the area is below the \(x\) - axis in the sense of the sign of \(f^{\prime}(x)\)), \(\int_{1}^{3}f^{\prime}(t)dt = 6\), \(\int_{3}^{5}f^{\prime}(t)dt=-3\), \(\int_{5}^{8}f^{\prime}(t)dt = 7\).
\(f(1)=f(0)+\int_{0}^{1}f^{\prime}(t)dt\), \(f(3)=f(1)+\int_{1}^{3}f^{\prime}(t)dt=f(0)+\int_{0}^{1}f^{\prime}(t)dt+\int_{1}^{3}f^{\prime}(t)dt=f(0)-2 + 6=f(0)+4\), \(f(5)=f(3)+\int_{3}^{5}f^{\prime}(t)dt=f(0)+4-3=f(0)+1\), \(f(8)=f(5)+\int_{5}^{8}f^{\prime}(t)dt=f(0)+1 + 7=f(0)+8\).
Since \(f(8) = 4\), then \(f(0)=-4\).
\(f(1)=-4-2=-6\), \(f(3)=-4 + 4 = 0\), \(f(5)=-4+1=-3\)

Step1: Recall the concavity and increasing - decreasing rules

The function \(y = f(x)\) is increasing when \(f^{\prime}(x)>0\) and concave down when \(f^{\prime\prime}(x)<0\). Since \(f^{\prime\prime}(x)\) is the derivative of \(f^{\prime}(x)\), \(f^{\prime\prime}(x)<0\) when \(f^{\prime}(x)\) is decreasing.
\(f^{\prime}(x)>0\) on the intervals \((1,3)\) and \((5,8)\).
\(f^{\prime}(x)\) is decreasing on the intervals \((0,1)\) and \((3,5)\).

Answer:

\(x = 1\) and \(x = 5\) are the values of \(x\) in the open interval \((0,8)\) for which \(f(x)\) has a local minimum.

(b)