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 $f$.

Short answer

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

Step-by-step solution

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

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

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