Find $f$ if $f''(x) = 3 \sqrt{x}$ if $f(4) = 20$, $f'(4) = 7$

Question

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 $'if'$ $f(4) = 20$, $f'(4) = 7$

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Is my math incorrect? Have I made any errors? Because the answer does not match.

Answer 1

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 $f(x).$

Answer 2

A problem like this can be classified into general diffirential equation or particular diffirential equation since he gave you 2 conditions this mean it's particular DE So all you have to do is to do the first integration and bring first constant from the condition he gave you then integrate again and bring the second constant from the second condition . You ignored C and C1 and that's wrong this is not limited integration