Find $f$ if $f''(x) = 3 \sqrt{x}$ if $f(4) = 20$, $f'(4) = 7$
$$\int \:\:3 \sqrt{x}\:dx = 2x^{\frac{3}{2}}+C \text{ (1st derivative) }$$
$$\int \:\: 2x^{\frac{3}{2}} dx = \dfrac{4x^\frac{5}{2}}{5} \text{ (original function) }$$
I think I did the integration right, but when I plug in $f(4)$, I get $25.6$, and when I plug in $f'(4)$, I get $16$. However the question says $'if'$ $f(4) = 20$, $f'(4) = 7$
Is my math incorrect? Have I made any errors? Because the answer does not match.
Best Answer
Hint
You have that $$f'(x)-f'(4)=\int_4^x f''(t)dt=\left[2t^{3/2}\right]_4^x=2x^{3/2}-2\cdot4^{3/2}=2x^{3/2}-4\sqrt 2.$$ Since $f'(4)=7$ we have
$$f'(x)=2x^{3/2}+7-4\sqrt 2.$$
Proceed in similar way to get $f(x).$