Question

first derivative test: problem 1
(1 point)
the function
$f(x)=-2x^{3}+12x^{2}+72x + 6$
is increasing on the interval
it is decreasing on the interval $(-\infty)$ and the interval $(\infty)$.
the function has a local maximum at
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Explanation:

Step1: Find the first derivative

Using the power rule \((x^n)^\prime=nx^{n - 1}\), for \(f(x)=-2x^{3}+12x^{2}+72x + 6\), we have \(f^\prime(x)=-6x^{2}+24x + 72\).
Factor out \(-6\): \(f^\prime(x)=-6(x^{2}-4x - 12)\).
Factor the quadratic: \(f^\prime(x)=-6(x - 6)(x+2)\).

Step2: Find critical points

Set \(f^\prime(x) = 0\), then \(-6(x - 6)(x + 2)=0\).
Solving \((x - 6)(x + 2)=0\) gives \(x=-2\) and \(x = 6\).

Step3: Determine intervals of increase and decrease

Use test - points in the intervals \((-\infty,-2)\), \((-2,6)\), and \((6,\infty)\).

• For \(x=-3\) (in \((-\infty,-2)\)): \(f^\prime(-3)=-6(-3 - 6)(-3 + 2)=-6\times(-9)\times(-1)=-54<0\). So \(f(x)\) is decreasing on \((-\infty,-2)\).
• For \(x = 0\) (in \((-2,6)\)): \(f^\prime(0)=-6(0 - 6)(0 + 2)=-6\times(-6)\times2 = 72>0\). So \(f(x)\) is increasing on \((-2,6)\).
• For \(x=7\) (in \((6,\infty)\)): \(f^\prime(7)=-6(7 - 6)(7 + 2)=-6\times1\times9=-54<0\). So \(f(x)\) is decreasing on \((6,\infty)\).

Step4: Find local maximum

By the first - derivative test, since \(f(x)\) changes from increasing to decreasing at \(x = 6\), the function has a local maximum at \(x = 6\).

Answer:

It is increasing on the interval \((-2,6)\). It is decreasing on the interval \((-\infty,-2)\) and the interval \((6,\infty)\). The function has a local maximum at \(6\).