# Given g(x) = sqrt(5x - 4) and h(x) = 4x^2 + 7 how do you find h(g(1))?

Mar 13, 2016

You must plug 1 in for x inside its respective function. Once you get the result from that calculation you must plug it into the adjacent function.

#### Explanation:

$h \left(g \left(1\right)\right) = 4 {\left(\sqrt{5 \left(1\right) - 4}\right)}^{2} + 7$

$h \left(g \left(1\right)\right) = 4 {\left(1\right)}^{2} + 7$

$h \left(g \left(1\right)\right) = 4 + 7$

$h \left(g \left(1\right)\right) = 11$

Don't forget to work from inside to the outside. Example: $f \left(g \left(h \left(x\right)\right)\right) =$h inside g inside f

Practice exercises:

1. Evaluate the following compositions if f(x) = 2^(x/2), g(x) = 3x^2 - 4x + 1 and h(x)=sqrt(2x + 9

a). $h \left(g \left(f \left(x\right)\right)\right)$

b). $g \left(f \left(g \left(h \left(x\right)\right)\right)\right)$

c). $g \left(h \left(f \left(6\right)\right)\right)$

Good luck!

Mar 13, 2016

$h \left(g \left(1\right)\right) = 11$

#### Explanation:

Doing one step at a time!

$\textcolor{b l u e}{\text{Consider g(1)}}$

$g \left(x\right) = \sqrt{5 x - 4}$

$\textcolor{b l u e}{\textcolor{m a \ge n t a}{g \left(1\right)} = \sqrt{5 \left(1\right) - 4} = \sqrt{5 - 4} = \sqrt{1} = \textcolor{red}{\pm 1}}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Consider "h(g(1))

$h \left(x\right) = 4 {x}^{2} + 7$

$h \left(g \left(x\right)\right) = 4 {\left(g \left(x\right)\right)}^{2} + 7$

$h \left(g \left(1\right)\right) = 4 {\left(\textcolor{m a \ge n t a}{g \left(1\right)}\right)}^{2} + 7$

$h \left(g \left(1\right)\right) = 4 {\left(\textcolor{red}{\pm 1}\right)}^{2} + 7$

But ${\left(- 1\right)}^{2} = {\left(+ 1\right)}^{2} = \textcolor{g r e e n}{+ 1}$

$h \left(g \left(1\right)\right) = 4 {\left(\textcolor{g r e e n}{+ 1}\right)}^{2} + 7$

$h \left(g \left(1\right)\right) = 4 + 7 = 11$