QuestionIf $\mathrm{H}=f \circ g$ and $H(x)=e^{4 x-5}+\frac{8 x-10}{7}$, then $g$ and f could be A $g(x)=4 x-5$ and $f(x)=e^{x}+\frac{x}{7}$ (B) $g(x)=e^{x}$ and $f(x)=x+\frac{2 x}{7}$ (C) $g(x)=4 x-5$ and $f(x)=e^{x}+\frac{2 x}{7}$ (D) $g(x)=8 x-10$ and $f(x)=e^{x}+\frac{x}{7}$

Studdy Solution
Check if $g(x) = 4x-5$ and $f(x) = e^{x} + \frac{2x}{7}$ satisfies $H(x) = f(g(x))$.
Calculate $f(g(x)) = f(4x-5) = e^{4x-5} + \frac{2(4x-5)}{7} = e^{4x-5} + \frac{8x-10}{7}$.
This matches $H(x)$.
The correct pair of functions is:
$$\boxed{(C) \, g(x) = 4x-5 \, \text{and} \, f(x) = e^{x} + \frac{2x}{7}}$$