Evaluate h(10), where h = (g ∘ f). f(x) = root3(x^2 - 4) g(x) = 14 x^3 + 4

Question

Hello, this is one of my homework questions but I'm very bad with square roots and have no clue on how to solve this

Answer 1

Hi,

When you evaluate one function "inside" another you read it f(g(x)) or f of g of x.

That little circle in the center is the same.  In this problem you want to find the opposite of what I wrote above.  You want to find "g of f" 

So....

f(x) = cube root of (x^2 - 4)  For clarity let's rename it.  Let f(x)=y

you want the g function of that.  So, g(f(x)) is now renamed g(y). Please note, the problem calls this "h."

Seems just a bit simpler....

SO....Where "x" is in the g(x) equation, you will substitute that/those x(s) for y..... and you will have h (or g(f(x))

so g(x) is above, substituting y for the only x, you get

h = g(f(x))= 14(root3(x^2-4))^3 +4 

Look at the center part, starting with the cube root, ending with the cube....

(root3(x^2-4))^3

If you cube something, and then take the cube root (I will show it next in exponents) you get the same something back!!

Example:  cube3(4)^3  Is really (4^3)^(1/3)

Sooooooo........you have 14(cube root ((something)^3) +4

The cube root and the cube cancel out.  The something was x^2-4

So you have: 14(x^2-4)+14  You can simplify from here!!!

Answer 2

f(x) = ³√(x²-4) =

 

g(x) = 14x³ + 4

 

h = g(f(x)) = 14(³√(x2-4))³ + 4

                 = 14(x²-4) + 4

 

h(10) = 14(10²-4) + 4

         = 14(96) + 4

         = 1348