Answer:
Function I
Step-by-step explanation:
Differentiate each function.
Function I
[tex]f(x)=e^{4-x}[/tex]
[tex]\boxed{\begin{minipage}{5.5 cm}\underline{Differentiating $e^{f(x)}$}\\\\If $y=e^{f(x)}$, then $\dfrac{\text{d}y}{\text{d}x}=f\:'(x)e^{f(x)}$\\\end{minipage}}[/tex]
[tex]f'(x)=-e^{4-x}[/tex]
The domain of f'(x) is all real numbers: (-∞, ∞)
Therefore f(x) is differentiable for all values of x over the interval (-5, 5).
Function II
[tex]f(x)=|4-x|[/tex]
[tex]\boxed{\begin{minipage}{6.5 cm}\underline{Differentiating an absolute value function}\\\\$|f(x)|'=\dfrac{f(x)}{|f(x)|}f'(x)$\\\end{minipage}}[/tex]
[tex]f'(x)=-\dfrac{4-x}{|4-x|}[/tex]
The domain of f'(x) is all real numbers except x = 4: (-∞, 4) ∪ (4, ∞)
Therefore, f(x) is not differentiable for all values of x over the interval (-5, 5).
Function III
[tex]\begin{aligned}f(x)&=\sqrt[3]{4-x}\\&=(4-x)^{\frac{1}{3}}\end{aligned}[/tex]
[tex]\boxed{\begin{minipage}{7 cm}\underline{Differentiating $[f(x)]^n$}\\\\If $y=[f(x)]^n$, then $\dfrac{\text{d}y}{\text{d}x}=n[f(x)]^{n-1} f'(x)$\\\end{minipage}}[/tex]
Therefore:
[tex]\begin{aligned}f'(x)&=\dfrac{1}{3}(4-x)^{\frac{1}{3}-1} \cdot -1\\&=-\dfrac{1}{3}(4-x)^{-\frac{2}{3}}\\&=-\dfrac{1}{3(4-x)^{\frac{2}{3}}}\end{aligned}[/tex]
The domain of f'(x) is all real numbers except x = 4: (-∞, 4) ∪ (4, ∞)
Therefore, f(x) is not differentiable for all values of x over the interval (-5, 5).
Function 4
[tex]f(x)=(4-x)^{\frac{2}{3}}[/tex]
[tex]\boxed{\begin{minipage}{7 cm}\underline{Differentiating $[f(x)]^n$}\\\\If $y=[f(x)]^n$, then $\dfrac{\text{d}y}{\text{d}x}=n[f(x)]^{n-1} f'(x)$\\\end{minipage}}[/tex]
Therefore:
[tex]\begin{aligned}f'(x)&=\dfrac{2}{3}(4-x)^{\frac{2}{3}-1} \cdot -1\\&=-\dfrac{2}{3}(4-x)^{-\frac{1}{3}}\\&=-\dfrac{2}{3(4-x)^{\frac{1}{3}}}\end{aligned}[/tex]
The domain of f'(x) is all real numbers except x = 4: (-∞, 4) ∪ (4, ∞)
Therefore, f(x) is not differentiable for all values of x over the interval (-5, 5).
