Given f(x) = 4x - 7, g(x) = 8x^2 - 2x - 3 and h(x) = 4x^3 + 9x - 2, find (g of f)(x), (h of f)(2) and (f of h)(x).

Step-by-step solution

1. 1

   Fix the convention. $(g \circ f)(x)$ means $g(f(x))$: the right function acts first. Reversing the order gives a different answer in general, and here $(f \circ g)(x) = 32x^2 - 8x - 19$, nothing like part (a).

2. 2

   Part (a): substitute f into g. Replace every $x$ in $g$ by $4x-7$:

   $g(f(x)) = 8(4x-7)^2 - 2(4x-7) - 3$

   With $(4x-7)^2 = 16x^2 - 56x + 49$:

   $= 128x^2 - 448x + 392 - 8x + 14 - 3 = 128x^2 - 456x + 403$

3. 3

   Part (b): evaluate inside first. For a numerical composition there is no need to build the general formula:

   $f(2) = 4(2) - 7 = 1, \qquad h(1) = 4(1)^3 + 9(1) - 2 = 11$

   so $(h \circ f)(2) = 11$. Composing symbolically first would mean expanding $(4x-7)^3$ for no gain.

4. 4

   Part (c): substitute h into f. Because $f$ is linear, this composition is easy — just scale and shift:

   $f(h(x)) = 4\left(4x^3 + 9x - 2\right) - 7 = 16x^3 + 36x - 8 - 7 = 16x^3 + 36x - 15$

5. 5

   Check the answers numerically. At $x = 1.7$: $f(1.7) = -0.2$ and $g(-0.2) = 0.32 + 0.4 - 3 = -2.28$, while the formula gives $128(2.89) - 456(1.7) + 403 = -2.28$. For (c) at $x = -2.3$: $h(-2.3) = -71.368$ and $f(-71.368) = -292.472$, while $16(-12.167) + 36(-2.3) - 15 = -292.472$. Both match.

6. 6

   Note the degree pattern. Composing a degree-$m$ with a degree-$n$ polynomial gives degree $mn$: $g \circ f$ is $2 \times 1 = 2$, and $f \circ h$ is $1 \times 3 = 3$, matching the results above.