determine any relative maxima or minima of the function and the intervals on which the function is…

determine any relative maxima or minima of the function and the intervals on which the function is increasing or decreasing\n$f(x)=-x^{2}+13x + 37$\ndoes the function have a relative maximum or minimum?\nrelative maximum\nrelative minimum\nthe relative maximum occurs at $x = square$ and has a value of $square$\non what interval is the function increasing?\na. $(6.25,infty)$\nb. $(-infty,5.25)$\nc. $(-infty,6.5)$\nd. $(6.5,infty)$\non what interval is the function decreasing?\na. $(-infty,6.5)$\nb. $(-infty,5.25)$
Answer
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): [ \begin{align*} -2x&=-13\ x&=\frac{13}{2}=6.5 \end{align*} ]
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): [ \begin{align*} f(6.5)&=-(6.5)^{2}+13\times6.5 + 37\ &=-42.25+84.5+37\ &=79.25 \end{align*} ]
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)).