Q. 1 - Calculus

Question

Fill in the blanks to complete each of the following theorem statements:

•  If $\mathrm{x} = \mathrm{c}$ is a local extremum of $f$, then $f\text{'}\left(c\right)$ is either $\_\_\_\_\_\_\_$or$\_\_\_\_\_\_\_$.

Step-by-Step Solution

Answer

Ans: part (a). $f_{+}^{\text{'}}\left(c\right)=\underset{x\to c^{+}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}\le 0$

        part (b). $f_{-}^{\text{'}}\left(c\right)=\underset{x\to c^{-}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}\ge 0$

Step 1. Given Information:

For a function $f\left(x\right)$ the local extremum is $x=c$ 

Step 2. Explanation:

• Suppose $x = c$ is the location of a local maximum of $f$.
• If $f\text{'}\left(c\right)$ does not exist, then $x=c$ is a critical point .
• And if $f\text{'}\left(c\right)$ exists, then it must be equal to 0.

$$\begin{gathered} Since x = c is the location of a local maximum \\ \delta > 0 \\ \therefore x \in (c-\delta , c+\delta ) \boxed{where f\left(c\right) \ge f\left(x\right)} \\ thus, f \left(x\right)-f \left(c\right) \le 0 \\ In the case where x \in (c, c+\delta ) ......[so x>c] \\ \therefore x - c is positive \\ case \frac{f \left(x\right)-f \left(c\right)}{x-c} \le 0 \\ {f\text{'}}_{+}\left(c\right) =\underset{x\to c^{+}}{\lim } \frac{f \left(x\right) - f \left(c\right)}{x - c} \le 0 \\ similarly x \in (c - \delta , c)......[so x<c] \\ and f \left(x\right) - f \left(c\right) \le 0, \\ {f\text{'}}_{-}\left(c\right) =\underset{x\to c^{-}}{\lim } \frac{f \left(x\right) - f \left(c\right)}{x - c} \ge 0 \end{gathered}$$