Question

find all relative extrema of the function. use the second derivative test where applicable. (if an answer does not exist, enter dne.)

$f(x)=14x - x^{2}$

relative maximum $(x,y)=$

relative minimum $(x,y)=$

Explanation:

Step1: Find the first derivative

The function is \(f(x)=14x - x^{2}\). Using the power rule \((x^n)^\prime=nx^{n - 1}\), the first derivative \(f^\prime(x)=(14x - x^{2})^\prime=14-2x\).

Step2: Find the critical points

Set \(f^\prime(x) = 0\), so \(14-2x=0\). Solving for \(x\):

$$
LATEXBLOCK0
$$

Step3: Find the second derivative

Differentiate \(f^\prime(x)=14 - 2x\) with respect to \(x\). Using the power rule, \(f^{\prime\prime}(x)=(14-2x)^\prime=- 2\).

Step4: Apply the second - derivative test

Since \(f^{\prime\prime}(7)=-2<0\), the function \(f(x)\) has a relative maximum at \(x = 7\).
Substitute \(x = 7\) into the original function \(f(x)=14x - x^{2}\): \(f(7)=14\times7-7^{2}=98 - 49=49\).

Since \(f^{\prime\prime}(x)=-2\) is always negative (there is no value of \(x\) for which \(f^{\prime\prime}(x)>0\)), there is no relative minimum.

Answer:

relative maximum \((x,y)=(7,49)\)
relative minimum \((x,y)=\text{DNE}\)