Hi, I\'m having trouble with this problem, could you help me out? All work must

Question

Hi, I'm having trouble with this problem, could you help me out? All work must be shown to help me understand what's going on.

A. Determine the domain of h(x).

B. Find h'(5/2)

C. At what x is h(x) a maximum? Show all the analysis that leads to your conclusion.

11. Let f be the function shown below, with domain the closed interval [0,6]. The graph of f shown below is composed of two semicircles (assume they touch). Let h(x) = integral 0 to 2x - 1 f(t)dt A. Determine the domain of h(x). B. Find h'(5/2) C. At what x is h(x) a maximum? Show all the analysis that leads to your conclusion.

Explanation / Answer

Part A. For defining h(x) ,the integral must be defined ,so for the intergral to be valid , it's limits must be proper .

since f(x) has domain [0,6] ,so the integral is valid on [0,6]

this implies 0<= 2x-1 <= 6 , i.e., 1/2 <= x <=7/2

So the domain of h(x) is [1/2 , 7/2] .

Part B. By fundamental theorem of calculus , h'(x)=d[h(x)]/dx = { d(2x-1)/dx }*f(2x-1) = 2f(2x-1)

h'(5/2) = 2*f(5-1) = 2 f(4) = 0

Part C. Theoritically , h(x) is just the area under graph f(t) vs t , and we know t varies from t=0 to t=6 .

So from t=0 to t=4 , area = 2 ? ,

and from t=4 to t=6 , area = -?/2

therefore , by visualisation , you can see area is maximum at t=4 on f(t) graph .

2x-1=t=4 , therefore, x=5/2

Also , h(0)=0 , h(3/2)=? , h (5/2) = 2? , h(3)=7?/4 , h(7/2)=3?/2

   So , h(x) is maximum at x=5/2 and h(x)max=2? .

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