Question

Consider the function f(x) whose second derivative is f''(x)= 8x+4sin(x), if f(0)=3 and f'(0)= 4, what is f(3)?

Answer 1

Answer:

Please find the attached image.

Step-by-step explanation:

Answer 2

integration of 2nd derivative gives first derivative with a constant,

Let see process !

$\int\limits^x_0{f"(x)}\,dx=\int\limits^x_0{(8x+4sinx)}\,dx$

or, $[f'(x)]^x_0=\left[4x^2-4cosx\right]^x_0$

or, f'(x) - f'(0) = 4x² - 4cosx + 4

given, f'(0) = 4

or, f'(x) - 4 = 4x² - 4cosx + 4

or, f'(x) = 4x² - cosx + 8

again, integrate f'(x) , we will get original function.

$\int\limits^x_0{f'(x)}\,dx=\int\limits^x_0{(4x^2-4cosx+8)}\,dx$

or, $[f(x)]^x_0=\left[\frac{4}{3}x^3-4sinx+8x\right]^x_0$

or, f(x) - f(0) = 4/3 x³ - 4sinx + 8x

given, f(0) = 3

so, f(x) - 3 = 4/3 x³ - 4sinx + 8x

or, f(x) = 4/3 x³ - 4sinx + 8x + 3

now, putting x = 3 to get f(3)

f(3) = 4/3 × 3³ - sin(3) + 8 × 3 + 3

= 36 - 4sin(3) + 24 + 3

= 63 - 4sin(3)

hence, f(3) = 63 - 4sin(3)