# Determine the value of f(ln2) when `f'(x) = e^x+4e^(-x) ` , and `f(0)=-3` . Find f(x) on (−pi/2,pi/2) when f'(x) = 5 + tan^(2) x and f(0) = 3.

First, determine the function f(x). To do so, integrate f'(x).

`f(t) = int f'(x) dx = int (e^x + 4e^(-x))dx = int e^x dx + 4 int e^(-x)dx`

Apply the formula of integral which is `int e^u du = e^u + C ` .

`f(t)= e^x + (- 4e^(-x) )+ C`

`f(t) = e^x - 4e^(-x) + C`

To determine the value of C, substitute `f(0)=-3` .

`-3=e^0 - 4e^0 + C`

Note that `e^0=1` .

`-3=1-4+C`

`-3=-3+C`

`0=C`

Hence, `f(x) = e^x-4e^(-x)` .

Next, to solve for f(ln2), replace the x in function f(x) with ln 2.

`f(x) = e^x - 4e^(-x)`

`f(ln2)=e^(ln2) - 4e^(-ln2)`

Apply the negative exponent rule which is `a^-m = 1/a^m` .

`f(ln2)=e^(ln2)-4/e^(ln2)`

Then, apply the rule of logarithm which is `e^(ln x) = x` . So, `e^(ln2) = 2` .

`f(ln2)=2-4/2`

`f(ln2)=2-2`

`f(ln2)=0`

**Hence, `f(ln2) = 0` .**

You need to use the inverse operation of differentiation to evaluate `f(x), ` hence, you need to integrate `f'(x)` such that:

`int f'(x) dx= int (5 + tan^2 x) dx = int (4 + 1 + tan^2 x)dx`

Using the linearity of integral yields:

`int f'(x) dx = int4 dx + int (1 + tan^2 x) dx `

You need to use the trigonometric identity such that:

`1 + tan^2 x = 1/(cos^2 x) `

`int f'(x) dx = int 4 dx + int 1/(cos^2 x) dx`

`int f'(x) dx = 4x + tan x + c`

Hence, evaluating f(x) yields:

`f(x) = 4x + tan x + c`

You need to find the constant c using the information provided by the problem f(0)=3 such that:

`f(0) = 4*0 + tan0 + c => c=3`

**Hence, evaluating the given function using the information provided by the problem, yields `f(x) = 4x + tan x + 3` .**