Question

Find f.f '(x) = square root( x) (6 + 5x), f(1) = 7

Answer 1

Since we have the first derivative of function f, then, we can find f as follows:

$f(x)=\int f^{\prime}(x)dx$

We can rewrite the given expression as follows

$f^{\prime}(x)=6x^{\frac{1}{2}}+5x^{\frac{3}{2}}$

because

$6\sqrt{x}=6x^{\frac{1}{2}}$

and

$5x\sqrt{x}=5x^{\frac{3}{2}}$

So, we need to compute

$f(x)=\int(6x^{\frac{1}{2}}+5x^{\frac{3}{2}})dx$

From the integration formula:

$\int x^ndx=\frac{x^{n+1}}{n+1}$

we get

$f(x)=6\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+5\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}+C$

where C is the constant of integration. From this result, we have

$f(x)=6\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+5\frac{x^{\frac{5}{2}}}{\frac{5}{2}}+C$

which gives

$f(x)=4x^{\frac{3}{2}}+2x^{\frac{5}{2}}+C$

or equivalently,

$f(x)=4\sqrt{x^3}+2\sqrt{x^5}+C$

Finally, we can find C by substituting the given information about f(x), that is, f(1)=7. It yields,

$7=4\sqrt{1^3}+2\sqrt{1^5}+C$

which gives

$\begin{gathered} 7=4+2+C \\ 7=6+C \end{gathered}$

Then

$C=1$

Therefore, the answer is:

$f(x)=4\sqrt{x^3}+2\sqrt{x^5}+1$