Q. 8 - Calculus

Question

The function $f\left(x\right)=4-x^2$ is both continuous and differentiable at x = 1. Write these facts as limit statements.

Step-by-Step Solution

Answer

The function is continuous and differentiable at $x=1$.

Step 1. Given information

We have to prove that the function $f\left(x\right)=4-x^2$ is both continuous and differentiable at $x=1$.

Step 2. Check the function is continuous

Left hand limit,

$$\begin{gathered} \underset{x\to 1^{-}}{\lim }4-x^2 \\ =4-1^2 \\ =3 \end{gathered}$$

Right hand limit,

$$\begin{gathered} \underset{x\to 1^{+}}{\lim }4-x^2 \\ =4-1^2 \\ =3 \end{gathered}$$

At point $x=1$, the value of function is

$$\begin{gathered} f\left(1\right)=4-1^2 \\ =3 \end{gathered}$$

Therefore, the function is continuous at $x=1$.

Step 3. Check the function is differentiable.

Left hand derivative,

$$\begin{gathered} \underset{h\to 0^{-}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h} \\ =\underset{h\to 0^{-}}{\lim }\frac{4-{\left(1+h\right)}^2-\left[4-1\right]}{h} \\ =\underset{h\to 0^{-}}{\lim }\frac{-2h-h^2}{h} \\ =\underset{h\to 0^{-}}{\lim }-2-h \\ =-2 \end{gathered}$$

Right hand derivative,

$$\begin{gathered} \underset{h\to 0^{+}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h} \\ =\underset{h\to 0^{+}}{\lim }\frac{4-{\left(1+h\right)}^2-\left[4-1\right]}{h} \\ =\underset{h\to 0^{+}}{\lim }\frac{-2h-h^2}{h} \\ =\underset{h\to 0^{+}}{\lim }-2-h \\ =-2 \end{gathered}$$

Therefore, the function is differentiable at $x=1$.