Given f' '(x) = - 4 sin(2x) and f'(0) = - 6 and f(0) = 4. Find

Question

Given  and  and .

Find 

Answer 1

Since we know the second derivative f''(x) = -4sin(2x), we can calculate the first derivative up to an unknown constant by taking the integral (anti-derivative) of f''(x).

f'(x) = ∫f''(x)dx = 2cos(2x) + C

Since we know the value of the first derivative (f'(x)) at x = 0, we can figure out what the unknown constant C is:

f'(0) = -6 = 2cos(0) + C → C = -6 - 2 = -8

Now we can write the first derivative as f'(x) = 2cos(2x) - 8.

We can do the same procedure to find a function for f(x). Let's integrate the first derivative f'(x)

f(x) = ∫f'(x)dx = sin(2x) + D

Let's plug in the known values that we have for f(x) to find the new unknown constant D

f(0) = 4 = sin(0) + D → D = 4

We can now write f(x) = sin(2x) + 4

Finally, we can evaluate f(π/5) = sin(2π/5) + 4 = 4.95