the figure above shows the graph of ( f ), the derivative of a twice - differentiable function ( f ), on the…

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 ).\n(a) find all values of ( x ) on the open interval ( 0lt xlt8 ) for which the function ( f ) has a local minimum. justify your answer.\n(b) determine the absolute minimum value of ( f ) on the closed interval ( 0leq xleq8 ). justify your answer.\n(c) on what open intervals contained in ( 0lt xlt8 ) is the graph of ( f ) both concave down and increasing? explain your reasoning.\n(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 ).

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 ).\n(a) find all values of ( x ) on the open interval ( 0lt xlt8 ) for which the function ( f ) has a local minimum. justify your answer.\n(b) determine the absolute minimum value of ( f ) on the closed interval ( 0leq xleq8 ). justify your answer.\n(c) on what open intervals contained in ( 0lt xlt8 ) is the graph of ( f ) both concave down and increasing? explain your reasoning.\n(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 ).

Answer

(a)

Explanation:

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 (1<x<3)). For (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 (1<x<3)), and to the right of (x = 3) (in an open interval around (x = 3)), (f^{\prime}(x)<0) (since (y = f^{\prime}(x)) is below the (x) - axis for (3<x<5)). For (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<x<5)), and to the right of (x = 5) (in an open interval around (x = 5)), (f^{\prime}(x)>0) (since (y = f^{\prime}(x)) is above the (x) - axis for (5<x<8)).

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)

Explanation:

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)

Answer:

The absolute minimum value of (f(x)) on the closed interval ([0,8]) is (-6) at (x = 1).

(c)

Explanation:

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:

The function (y = f(x)) is concave down and increasing on the interval ((5,8)).

(d)

Explanation:

Step1: Use the chain rule

If (g(x)=(f(x))^{3}), then by the chain rule (g^{\prime}(x)=3(f(x))^{2}\cdot f^{\prime}(x)). We need to find (f^{\prime}(3)). The slope of the tangent line to (y = f^{\prime}(x)) at (x = 3) is (f^{\prime}(3)). From the graph of (y = f^{\prime}(x)), (f^{\prime}(3)=0) (horizontal tangent line at (x = 3)).

Step2: Calculate (g^{\prime}(3))

Substitute (x = 3) into (g^{\prime}(x)): (g^{\prime}(3)=3(f(3))^{2}\cdot f^{\prime}(3)). Since (f(3)=-\frac{5}{2}) and (f^{\prime}(3) = 0)

Answer:

(g^{\prime}(3)=0)