Question

determine any relative maxima or minima of the function and the intervals on which the function is increasing or decreasing
$f(x)=-x^{2}+13x + 37$
does the function have a relative maximum or minimum?
relative maximum
relative minimum
the relative maximum occurs at $x = square$ and has a value of $square$
on what interval is the function increasing?
a. $(6.25,infty)$
b. $(-infty,5.25)$
c. $(-infty,6.5)$
d. $(6.5,infty)$
on what interval is the function decreasing?
a. $(-infty,6.5)$
b. $(-infty,5.25)$

Explanation:

Step1: Find the derivative of the function

The function is \(f(x)=-x^{2}+13x + 37\). Using the power rule \((x^{n})^\prime=nx^{n - 1}\), the derivative \(f^\prime(x)=-2x + 13\).

Step2: Find the critical point

Set \(f^\prime(x)=0\), so \(-2x+13 = 0\). Solving for \(x\):

$$
LATEXBLOCK0
$$

Step3: Use the second - derivative test

The second - derivative \(f^{\prime\prime}(x)=-2\). Since \(f^{\prime\prime}(x)=-2<0\), the function has a relative maximum.

Step4: Find the value of the relative maximum

Substitute \(x = 6.5\) into the original function \(f(x)=-x^{2}+13x + 37\):

$$
LATEXBLOCK1
$$

Step5: Determine the intervals of increase and decrease

We know that \(f^\prime(x)=-2x + 13\).

• For the increasing interval: Set \(f^\prime(x)>0\), \(-2x + 13>0\), \(x<6.5\). So the function is increasing on the interval \((-\infty,6.5)\).
• For the decreasing interval: Set \(f^\prime(x)<0\), \(-2x + 13<0\), \(x>6.5\).

Answer:

• The function has a Relative maximum.
• The relative maximum occurs at \(x = 6.5\) and has a value of \(79.25\).
• The function is increasing on the interval \(C.(-\infty,6.5)\).
• The function is decreasing on the interval \(A.(6.5,\infty)\).