Question

consider the following function. use a graphing utility to confirm your answers for parts (a) through (c). (if an answer does not exist, enter dne.)

$f(x)=-4x^{2}+24x + 9$

(a) find the critical numbers of f. (enter your answers as a comma-separated list.)

$x=$

(b) find the open intervals on which the function is increasing or decreasing. (enter your answers using interval notation.)

increasing

decreasing

(c) apply the first derivative test to identify all relative extrema.

relative maximum $(x,y)=()$

relative minimum $(x,y)=()$

Explanation:

Step1: Find the derivative of \(f(x)\)

The function is \(f(x)=-4x^{2}+24x + 9\). Using the power rule \((x^{n})^\prime=nx^{n - 1}\), we have \(f^\prime(x)=-8x+24\).

Step2: Find the critical numbers

Set \(f^\prime(x) = 0\). So, \(-8x + 24=0\). Solving for \(x\):

$$
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Step3: Determine the intervals of increase and decrease

• For the interval \((-\infty,3)\), let's test \(x = 0\). Then \(f^\prime(0)=-8\times0 + 24=24>0\). So the function is increasing on \((-\infty,3)\).
• For the interval \((3,\infty)\), let's test \(x = 4\). Then \(f^\prime(4)=-8\times4+24=-8<0\). So the function is decreasing on \((3,\infty)\).

Step4: Apply the First - Derivative Test for relative extrema

Since \(f(x)\) changes from increasing \((f^\prime(x)>0)\) to decreasing \((f^\prime(x)<0)\) at \(x = 3\).
Find \(y\) - value: \(y=f(3)=-4\times3^{2}+24\times3 + 9=-36 + 72+9=45\). There is a relative maximum at \((3,45)\) and no relative minimum (because the function is a parabola opening downwards (\(a=-4<0\)) and has only one critical point).

Answer:

(a) \(x = 3\)
(b) Increasing: \((-\infty,3)\); Decreasing: \((3,\infty)\)
(c) Relative maximum: \((x,y)=(3,45)\); Relative minimum: DNE