Question – 2: Consider the function,f(x) = (2/x+2)+4exp(-2x).

(a) Find the area under the graph,A,and the average value,W, of the function,f(x),in the interval,[2,5].

(b) Find, f'(x), f''(x-2),and,f'(0) .

(c) If, g(x) = x^2-1,find f(g(x)),and
d/dx(f(g(x)) ?

(a) To find the area under the graph of a function within a given interval, you need to integrate the function over that interval. In this case, you want to find the area under the graph of f(x) in the interval [2, 5]. The integral of f(x) can be calculated as follows:

∫[2, 5] f(x) dx = ∫[2, 5] (2/x + 2) + 4exp(-2x) dx

To solve this integral, you can break it down into two separate integrals. First, integrate (2/x + 2) with respect to x:

∫(2/x + 2) dx = 2ln|x| + 2x + C

Next, integrate 4exp(-2x) with respect to x:

∫4exp(-2x) dx = -2exp(-2x) + C

Now, you can evaluate the definite integral over the interval [2, 5]:

∫[2, 5] f(x) dx = [2ln|x| + 2x + C - 2exp(-2x) + C] evaluated from x = 2 to x = 5

Plug in the values of x=5 and x=2:

A = [2ln|5| + 2(5) + C - 2exp(-2*5) + C] - [2ln|2| + 2(2) + C - 2exp(-2*2) + C]

Simplify further to get the numerical value of A.

To find the average value of the function f(x) in the interval [2, 5], you need to divide the area (A) by the width of the interval, (5-2), which is 3:

W = A / (5 - 2)

(b) To find the derivative f'(x) of the function f(x), you need to differentiate each term of the function separately.

f(x) = (2/x + 2) + 4exp(-2x)

f'(x) = (d/dx)(2/x) + (d/dx)(2) + (d/dx)(4exp(-2x))

Using the power rule, chain rule, and product rule, respectively, you can calculate the derivatives of each term:

f'(x) = (-2/x^2) + 0 + (-8exp(-2x))

To find f''(x-2), you use the chain rule. Let u = x - 2:

f''(x-2) = d^2/dx^2 [f(u)] = d/du [f'(u)]

First, differentiate f'(u) with respect to u:

f'(u) = -2/u^2 - 8exp(-2u)

Then, differentiate f'(u) with respect to u:

f''(x-2) = d/du [f'(u)] = -2/u^2 - 16exp(-2u)

Finally, computing f'(0) involves plugging in x=0 into f'(x):

f'(0) = (-2/0^2) + 0 + (-8exp(0))

Simplify further to get the numerical value of f'(0).

(c) To find f(g(x)), you need to substitute g(x) into f(x). In this case, g(x) = x^2 - 1:

f(g(x)) = f(x^2 - 1)

Substitute x^2 - 1 into f(x):

f(g(x)) = (2/(x^2 - 1) + 2) + 4exp(-2(x^2 - 1))

To find d/dx (f(g(x))), you need to differentiate f(g(x)) with respect to x:

d/dx (f(g(x))) = d/dx [(2/(x^2 - 1) + 2) + 4exp(-2(x^2 - 1))]

Using the chain rule and product rule, you can calculate the derivative:

d/dx (f(g(x))) = (-4x/(x^2 - 1)^2) + 8xexp(-2(x^2 - 1))