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

2 Answers
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(g(1)) = 4(sqrt(5(1) - 4))^2 + 7#

#h(g(1)) = 4(1)^2 + 7#

#h(g(1)) = 4 + 7#

#h(g(1)) = 11#

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

b). #g(f(g(h(x))))#

c). #g(h(f(6)))#

Good luck!

Mar 13, 2016

#h(g(1))=11#

Explanation:

Doing one step at a time!

#color(blue)("Consider g(1)")#

#g(x)=sqrt(5x-4)#

#color(blue)(color(magenta)(g(1))= sqrt(5(1)-4) = sqrt(5-4)=sqrt(1)=color(red)(+-1))#

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

#h(x)=4x^2+7#

#h(g(x))=4(g(x))^2+7#

#h(g(1))=4(color(magenta)(g(1)))^2+7#

#h(g(1))=4(color(red)(+-1))^2+7#

But #(-1)^2=(+1)^2 =color(green)( +1)#

#h(g(1))=4(color(green)(+1))^2+7#

#h(g(1))=4+7 =11#