Q. 7 - Calculus

Question

The function $f\left(x\right)=4x^3-5x+1$ is both continuous and differentiable at $x=2$. Write these facts as limit statements.

Step-by-Step Solution

Answer

The function is continuous and differentiable at $x=2$.

Step 1. Given information

We have to prove that the function $f\left(x\right)=4x^3-5x+1$ is both continuous and differentiable at $x=2$.

Step 2. Check the function is continuous

Left hand limit,

$$\begin{gathered} \underset{x\to 2^{-}}{\lim }4x^3-5x+1 \\ =4{\left(2\right)}^3-5\left(2\right)+1 \\ =23 \end{gathered}$$

Right hand limit,

$$\begin{gathered} \underset{x\to 2^{+}}{\lim }4x^3-5x+1 \\ =4{\left(2\right)}^3-5\left(2\right)+1 \\ =23 \end{gathered}$$

At point $x=2$, the value of function is

$$\begin{gathered} f\left(2\right)=4{\left(2\right)}^3-5\left(2\right)+1 \\ =23 \end{gathered}$$

Therefore, the function is continuous at $x=2$.

Step 3. Check the function is differentiable.

Left hand derivative,

$$\begin{gathered} \underset{h\to 0^{-}}{\lim }\frac{f\left(2+h\right)-f\left(2\right)}{h} \\ =\underset{h\to 0^{-}}{\lim }\frac{4{\left(2+h\right)}^3-5\left(2+h\right)+1-\left[4{\left(2\right)}^3-5\left(2\right)+1\right]}{h} \\ =\underset{h\to 0^{-}}{\lim }\frac{48h+24h^2+4h^3-5h}{h} \\ =\underset{h\to 0^{-}}{\lim }43+24h+4h^2 \\ =43 \end{gathered}$$

Right hand derivative,

$$\begin{gathered} \underset{h\to 0^{+}}{\lim }\frac{f\left(2+h\right)-f\left(2\right)}{h} \\ =\underset{h\to 0^{+}}{\lim }\frac{4{\left(2+h\right)}^3-5\left(2+h\right)+1-\left[4{\left(2\right)}^3-5\left(2\right)+1\right]}{h} \\ =\underset{h\to 0^{+}}{\lim }\frac{48h+24h^2+4h^3-5h}{h} \\ =\underset{h\to 0^{+}}{\lim }43+24h+4h^2 \\ =43 \end{gathered}$$

Therefore, the function is differentiable at $x=2$.