Calculus everywhere

Calculus Level 3

Let $f$ be a twice differentiable function such that $f''(x) = -f(x)$ and $f'(x) = g(x)$ . If $h'(x)= {f(x)}^2 + {g(x)}^2$ , $h(1)=8$ and $h(0)=2$ , then what is $h(2)=?$

The answer is 14.

1 solution

Ojasvi Sharma
Aug 25, 2017

h'(x) =[f(x)]^2+[g(x)]^2 Therefore, h"(x) = 2f(x)f'(x) +2g(x)g'(x) = 2f(x)g(x) +2g(x)f"(x). [Since f'(x) = g(x)] = 2f(x)g(x) - 2g(x)f(x). [Since f"(x)= - f(x)] =0 Thus h'(x)= k, a constant for all x belong to R. Hence h(x) =k+m. From h(0)=2, we get m=2 and from h(1)=8 we get k=6. Therefore h(2)=8+6=14.

How did you go from h'(x)=0 to h'(x)=k?

John Crocker - 3 years, 9 months ago

As h"(x)=0 therefore h'(x)=k ( a constant). Because differentiation of a constant is zero. Here if h(x) would be a constant,then only it's derivative h"(x) would be 0.

Ojasvi Sharma - 3 years, 9 months ago

I meant to say if h'(x) would be a constant, then only it's derivative h"(x) would be zero.

Ojasvi Sharma - 3 years, 9 months ago

Why is h'(x) = f(x)^2 + g(x)^2? The problem says that's h(x).

Peter van der Linden - 3 years, 9 months ago

Actually, you are right. I apologize for the typing mistake.

Ojasvi Sharma - 3 years, 9 months ago

Calculus everywhere

The answer is 14.

1 solution

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