Question

To find the familiar "second derivative test "for a maximum or minimum point. That is show that ${\mathit{f}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{=}\mathbf{0}$, then${\mathit{f}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{>}\mathbf{0}$implies a minimum point at$\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathit{a}$and${\mathit{f}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{<}\mathbf{0}$implies a maximum point at $\mathit{x}\mathbf{=}\mathit{a}$.

Short answer

For all $x$near enough to $a$is given as follows:

$$\begin{gathered} f\left(x\right)-f\left(a\right)>0 \\ f\left(x\right)>f\left(a\right) \end{gathered}$$

Step-by-step solution

Given data

The given function is $f^{\text{'}}\left(a\right)=0$, and then $f^{\text{'}\text{'}}\left(a\right)>0$which implies a minimum point at $x=a$ and $f^{\text{'}\text{'}}\left(a\right)<0$ implies a maximum point at $x=a$.

Concept of Taylor series

Taylor series representation off about$\mathit{x}\mathbf{=}\mathit{a}$is given as:

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{+}\frac{{\mathbf{f}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{a}\mathbf{\right)}}{\mathbf{1}\mathbf{!}}\mathbf{+}\frac{{\mathbf{f}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{\left(}\mathbf{a}\mathbf{\right)}}{\mathbf{2}\mathbf{!}}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{a}{\mathbf{)}}^{\mathbf{2}}$ …(1)

Calculation to find the maximum point for the derivative test f

Suppose that $f^{\text{'}}\left(a\right)=0$then from (1) obtain:

$$\begin{gathered} f^{\text{'}\text{'}}\left(a\right)>0 \\ 2\frac{f\left(x\right)-f\left(a\right)}{{(x-a)}^2}>0 \end{gathered}$$

For all $x$near enough to $a$is given as follows:

$$\begin{gathered} f\left(x\right)-f\left(a\right)>0 \\ f\left(x\right)>f\left(a\right) \end{gathered}$$

That is, $f$increases.

Now, by the first derivative test $f$has minimum point at $x=a$:

$$\begin{gathered} f^{\text{'}\text{'}}\left(a\right)<0 \\ \frac{2f\left(x\right)-f\left(a\right)}{(x-a)}<0 \\ f\left(x\right)-f\left(a\right)<0 \end{gathered}$$

$f\left(x\right)<f\left(a\right)$for all $x$near enough to $a$.

That is, $f$decreases.

Now, by the first derivative test $f$has maximum point at $x=a$.