Answer
a) The value of $\left( f\circ g \right)\left( x \right)$ is $-{{x}^{2}}-10x-21$.
b) The value of $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ is $-2x-h,\text{ }h\ne 0$.
Work Step by Step
(a)
Consider the functions, $f\left( x \right)=4-{{x}^{2}}$ and $g\left( x \right)=x+5$
The value of the function $\left( f\circ g \right)\left( x \right)$ is,
$\left( f\circ g \right)\left( x \right)=f\left( g\left( x \right) \right)$
Replace $x$ by $g\left( x \right)$ in the function $f\left( x \right)=4-{{x}^{2}}$.
$f\left( g\left( x \right) \right)=4-{{\left[ g\left( x \right) \right]}^{2}}$
Substitute $g\left( x \right)=x+5$ in the function $f\left( g\left( x \right) \right)=4-{{\left[ g\left( x \right) \right]}^{2}}$.
$f\left( g\left( x \right) \right)=4-{{\left( x+5 \right)}^{2}}$
Use the property ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$.
$\begin{align}
& f\left( g\left( x \right) \right)=4-\left( {{x}^{2}}+10x+25 \right) \\
& =4-{{x}^{2}}-10x-25 \\
& =-{{x}^{2}}-10x-21
\end{align}$
Therefore, the value of $\left( f\circ g \right)\left( x \right)$ is $-{{x}^{2}}-10x-21$.
(b)
Consider the function, $f\left( x \right)=4-{{x}^{2}}$
Substitute $x+h$ for $x$ in the function $f\left( x \right)=4-{{x}^{2}}$.
$f\left( x+h \right)=4-{{\left( x+h \right)}^{2}}$
Consider the expression, $\frac{f\left( x+h \right)-f\left( x \right)}{h}$
Substitute $f\left( x+h \right)=4-{{\left( x+h \right)}^{2}}$ and $f\left( x \right)=4-{{x}^{2}}$ in the expression $\frac{f\left( x+h \right)-f\left( x \right)}{h}$.
$\frac{f\left( x+h \right)-f\left( x \right)}{h}=\frac{\left[ 4-{{\left( x+h \right)}^{2}} \right]-\left( 4-{{x}^{2}} \right)}{h}$
Use the property ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$.
$\begin{align}
& \frac{f\left( x+h \right)-f\left( x \right)}{h}=\frac{\left[ 4-\left( {{x}^{2}}+2xh+{{h}^{2}} \right) \right]-4+{{x}^{2}}}{h} \\
& =\frac{4-{{x}^{2}}-2xh-{{h}^{2}}-4+{{x}^{2}}}{h} \\
& =\frac{-2xh-{{h}^{2}}}{h} \\
& =-2x-h
\end{align}$
Therefore, the value of $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ is $-2x-h,\text{ }h\ne 0$.
