Question

- 1 POINTS GHCOLALG11 3.6.042. Let f(x) = 2x - 5 and g(x) = 5x – 2. Find the value. (gog) (909({}) - Need Help? Read It Watch It Talk to a Tutor -/1 POINTS GHCOLALG11 3.6.044.MI. Let f(x) = 5x2 - 4 and g(x) = 6x + 6. Find the value. (go (2) (gon(2) =

Answer 1

f(1) = 2.1 – 5, g(x) = 5x – 2

.

$\left(g\:\circ \:g\right)\left(x\right)=g\left(g\left(x\right)\right)$

$\left(g\:\circ \:g\right)\left(x\right)=g\left(5x-2\right)$

gog)(x) = 5(5.x - 2) - 2

$\left(g\:\circ \:g\right)\left(x\right)=25x-10-2$

$\left(g\:\circ \:g\right)\left(x\right)=25x-12$

take x=3/5

$\left(g\:\circ \:g\right)\left( \frac{3}{5}\right)= 25\left(\frac{3}{5}\right)-12$

$\left(g\:\circ \:g\right)\left( \frac{3}{5}\right)= 15-12$

${\color{Red} \left(g\:\circ \:g\right)\left( \frac{3}{5}\right)= 3}$

.

.

$f\left(x\right)=5x^2-4,\:\:\:\:g\left(x\right)=6x+6$

$\left(g\:\circ \:f\right)\left(x\right)=g\left(f\left(x\right)\right)$

$\left(g\:\circ \:f\right)\left(x\right)=g\left(5x^2-4\right)$

$\left(g\:\circ \:f\right)\left(x\right)=6\left(5x^2-4\right)+6$

$\left(g\:\circ \:f\right)\left(x\right)=30x^2-24+6$

$\left(g\:\circ \:f\right)\left(x\right)=30x^2-18$

take x=2

$\left(g\:\circ \:f\right)\left(2\right)=30 (2)^2-18$

$\left(g\:\circ \:f\right)\left(2\right)=30 (4)-18$

$\left(g\:\circ \:f\right)\left(2\right)=120-18$

${\color{Red} \left(g\:\circ \:f\right)\left(2\right)= 102}$