Q. 6 - Calculus

Question

Suppose that $f$ is defined for $x=0$ and differentiable everywhere except at $x=0$ and $x=1$, and that $f\text{'}\left(x\right)=0$ only at $x$= ±2. List all the critical points of f and sketch a possible graph of $f$.

Step-by-Step Solution

Answer

$f\left(x\right)=0.2x^5-0.25x^4-1.33x^3-2x^2$

Step1. Given Information

$$\begin{gathered} \text{ The function is defined for }x\ne 0\text{ and is differentiable at all points except at }x=0\text{ and }x=1\text{ and that }{f}^{\mathrm{\prime }}\left(x\right)=0 \\ \text{ only at }x=\pm 2 \end{gathered}$$

Step 2. Calculations

$$\begin{gathered} \text{ The critical points of the function, therefore will be } \\ x=-2;x=0;x=1;x=2 \\ \text{ Therefore, the differentiation of the function will be } \\ {f}^{\mathrm{\prime }}\left(x\right)=x(x+2)(x-2)(x-1) \\ \text{ The graph can be drawn as } \end{gathered}$$

$$\begin{gathered} \text{ Therefore, the function is } \\ f\left(x\right)=0.2x^5-0.25x^4-1.33x^3-2x^2 \end{gathered}$$