# The curve with equation y = h(x) passes through the point (4, 19). Given that h'(x) = 15xsqrtx-40/sqrtx, find h(x)?

Mar 4, 2018

since $h ' \left(x\right) = 15 {x}^{\frac{3}{2}} - 40 {x}^{- \frac{1}{2}}$

the antiderivative or indefinite integral of $h ' \left(x\right)$ will be
$h \left(x\right)$
therefore
$\int 15 {x}^{\frac{3}{2}} - 40 {x}^{- \frac{1}{2}} \mathrm{dx}$

this can be integrated easily by using the inverse of the Power rule

and it comes out as

$\frac{2}{5} \cdot 15 {x}^{\frac{5}{3}} - 80 {x}^{\frac{1}{2}} + C$
$= 6 {x}^{\frac{5}{2}} - 80 {x}^{\frac{1}{2}} + C$

to find C, we can use the fact that $h \left(x\right)$ passes through $\left(4 , 19\right)$
so $h \left(4\right) = 19$

therefore
$6 \cdot {4}^{\frac{5}{2}} - 80 \cdot {4}^{\frac{1}{2}} + C = 19$

$6 \cdot 32 - 80 \cdot 2 + C = 19$

$192 - 160 + C = 19$

$C = 19 - 32$

$\therefore$

$C = - 13$

therefore, the entire function $h \left(x\right)$ is

# $h \left(x\right) = 6 {x}^{\frac{5}{2}} - 80 {x}^{\frac{1}{2}} - 13$

and you can factorise it more if you want