Given f ' ' ( x ) = − 25 sin ( 5 x ) and f ' ( 0 ) = − 5 and f ( 0 ) = 3 .

Question

Find f(π/2) =

Answer 1

f(x) = sin5x-10x +3

f(0)=sin0 -10(0)+3 = 3

f'(x)=5cos5x-10

f'(0)=5cos0 -10 =5(1)-10=-5

f"(x)=-25sin5x

f(pi/2)=sin(5pi/2)-10pi/2 + 3

=1 -10pi/2 + 3

=4-5pi

Answer 2

Hi Jay!

f''(x) = -25sin(5x)

First step is to find f'(x).

f'(x) = ∫f''(x)

f'(x) = ∫(-25sin(5x))dx du du

f'(x) = -25∫sin(5x)dx let u = 5x, then --- = 5 or ----- = dx

dx 5

du

f'(x) = -25∫sin(u) ---

5

f'(x) = -5∫sin(u)du

f'(x) = -5 [ -cos(u) ] + C

f'(x) = -5 [ -cos(5x)] + C

but f'(0) = -5, so

5cos(5*0) + C = -5

5(1) + C = -5

C = -10

f'(x) = 5cos(5x)-10

Now, find f(x)

f(x) = ∫f'(x)

f(x) = ∫[5cos(5x)-10]dx

f(x) = 5∫cos(5x)dx - 10∫dx

du du

let u = 5x, then ---- = 5 or ---- = dx

dx 5

du

f(x) = 5∫cos(u) --- - 10x + C

5

f(x) = ∫cos(u)du - 10x + C

f(x) = sin(u) - 10x + C

f(x) = sin(5x) - 10x + C

We're told f(0) = 3, so

sin(5*0) - 10(0) + C = 3

0 - 0 + C = 3

C = 3

f(x) = sin(5x) - 10x + 3

Now, to find f(π/2)

f(π/2) = sin(5*π/2) - 10(π/2) + 3

= 1 - 5π + 3

= 4 - 5π