SL 5.6—Stationary points, local max and min

$f^{\prime }(x)=2x-8$.

Solve $2x-8=0$.

$2x=8\Rightarrow x=4$.

The stationary point occurs at $x=4$.

$x=-4$ results from a sign error when solving $2x-8=0$.

$x=8$ comes from forgetting to divide by 2.

$x=3$ is the constant term of $f(x)$, not related to the derivative.

Only $x=4$ satisfies $f^{\prime }(x)=0$.

A stationary point requires $f^{\prime }(x)=0$.

Here $f^{\prime }(x)=5$ for all $x$, which is never zero.

Since $f^{\prime }(x)\ne 0$ anywhere, no stationary points exist.

The function is a straight line with constant positive gradient, always increasing.

A local maximum requires $f^{\prime }(x)=0$ with sign change from + to -, which never happens here.

A local minimum requires $f^{\prime }(x)=0$ with sign change from - to +, which cannot occur since $f^{\prime }$ is constant.

A horizontal point of inflection also requires $f^{\prime }(x)=0$ somewhere, which is not satisfied.

No stationary points exist since $f^{\prime }(x)$ is never zero.