Question

question 9 the function $f(x)=2x^{3}-42x^{2}+270x + 3$ has one local minimum and one local maximum. use a graph of the function to estimate these local extrema. this function has a local minimum at $x =$ with output value: and a local maximum at $x =$ with output value: question help: video message instructor post to forum

Explanation:

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

$f'(x)=6x^{2}-84x + 270$

Step2: Set the derivative equal to zero

$6x^{2}-84x + 270 = 0$
Divide through by 6: $x^{2}-14x + 45=0$
Factor: $(x - 5)(x - 9)=0$
Solve for \(x\): \(x = 5\) or \(x=9\)

Step3: Determine local - extrema type

Take the second - derivative \(f''(x)=12x-84\)
When \(x = 5\), \(f''(5)=12\times5-84=- 24<0\), so \(x = 5\) is a local maximum.
When \(x = 9\), \(f''(9)=12\times9-84 = 24>0\), so \(x = 9\) is a local minimum.

Step4: Find the output values

For \(x = 9\) (local minimum), \(f(9)=2\times9^{3}-42\times9^{2}+270\times9 + 3=2\times729-42\times81+2430 + 3=1458-3402+2430 + 3=489\)
For \(x = 5\) (local maximum), \(f(5)=2\times5^{3}-42\times5^{2}+270\times5 + 3=2\times125-42\times25+1350 + 3=250-1050+1350 + 3=553\)

Answer:

This function has a local minimum at \(x = 9\)
with output value: \(489\)
and a local maximum at \(x = 5\)
with output value: \(553\)