Let f(x)=2x+3 and g(x)=x^2-4 and h(x)=x-3/2, how do you find g(f(3))?

May 11, 2015

The answer is : $g \left(f \left(3\right)\right) = 77$

Firstly, let's find $g \left(f \left(x\right)\right)$ :

You have $f \left(x\right) = 2 x + 3$ and $g \left(x\right) = {x}^{2} - 4$, so

$g \left(f \left(x\right)\right) = {f}^{2} \left(x\right) - 4 = {\left(2 x + 3\right)}^{2} - 4$

$g \left(f \left(x\right)\right) = 4 {x}^{2} + 12 x + 9 - 4 = 4 {x}^{2} + 12 x + 5$

Now you can calculate $g \left(f \left(3\right)\right)$ :

$g \left(f \left(3\right)\right) = 4 \cdot {3}^{2} + 12 \cdot 3 + 5 = 36 + 36 + 5 = 77$

May 11, 2015

Have a look. I think it is ok, although I didn't used $h \left(x\right)$...