Question

the function ( s(x) = -2x^{3}+24x^{2}+54x + 10 ) has one local minimum and one local maximum. graph the function on your calculator to estimate these local extrema and the intervals on which ( s ) is increasing and decreasing.
( s(x) ) has a local maximum of ( ) at ( x = )
( s(x) ) has a local minimum of ( ) at ( x = ).
( s(x) ) is increasing on the interval(s): ( )
( s(x) ) is decreasing on the interval(s): ( )
question help: video

Explanation:

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

The derivative \(s^{\prime}(x)=-6x^{2}+48x + 54=-6(x^{2}-8x - 9)=-6(x + 1)(x - 9)\)

Step2: Find the critical points

Set \(s^{\prime}(x)=0\), then \(-6(x + 1)(x - 9)=0\), so \(x=-1\) or \(x = 9\)

Step3: Use the first - derivative test

• For \(x<-1\), let \(x=-2\), then \(s^{\prime}(-2)=-6(-2 + 1)(-2 - 9)=-6\times(-1)\times(-11)=-66<0\)
• For \(-1
• For \(x>9\), let \(x = 10\), then \(s^{\prime}(10)=-6(10 + 1)(10 - 9)=-6\times11\times1=-66<0\)

Since the function changes from decreasing (\(s^{\prime}(x)<0\)) to increasing (\(s^{\prime}(x)>0\)) at \(x=-1\), \(s(x)\) has a local minimum at \(x=-1\). \(s(-1)=-2(-1)^{3}+24(-1)^{2}+54(-1)+10=2 + 24-54 + 10=-18\)

Since the function changes from increasing (\(s^{\prime}(x)>0\)) to decreasing (\(s^{\prime}(x)<0\)) at \(x = 9\), \(s(x)\) has a local maximum at \(x = 9\). \(s(9)=-2(9)^{3}+24(9)^{2}+54(9)+10=-1458+1944+486 + 10=982\)

Answer:

\(s(x)\) has a local maximum of \(982\) at \(x = 9\)

\(s(x)\) has a local minimum of \(-18\) at \(x=-1\)

\(s(x)\) is increasing on the interval(s): \((-1,9)\)

\(s(x)\) is decreasing on the interval(s): \((-\infty,-1)\cup(9,\infty)\)