Question

the function ( f(x)=-2x^{3}+42x^{2}-270x + 5 ) has one local minimum and one local maximum. this function has a local minimum at ( x ) equals with value and a local maximum at ( x ) equals with value

Explanation:

Step1: Find the first - derivative

The function is \(f(x)=-2x^{3}+42x^{2}-270x + 5\).
Using the power rule \((x^{n})^\prime=nx^{n - 1}\), we get \(f^\prime(x)=-6x^{2}+84x - 270\).
Factor out \(-6\): \(f^\prime(x)=-6(x^{2}-14x + 45)\).
Factor the quadratic: \(f^\prime(x)=-6(x - 5)(x - 9)\).

Step2: Find the critical points

Set \(f^\prime(x)=0\), then \(-6(x - 5)(x - 9)=0\).
Solving \(x-5 = 0\) gives \(x = 5\), and solving \(x - 9=0\) gives \(x = 9\).

Step3: Use the second - derivative test

Find the second - derivative. \(f^\prime(x)=-6x^{2}+84x - 270\), so \(f^{\prime\prime}(x)=-12x + 84\).
When \(x = 5\), \(f^{\prime\prime}(5)=-12\times5+84=-60 + 84=24>0\).
When \(x = 9\), \(f^{\prime\prime}(9)=-12\times9+84=-108 + 84=-24<0\).
Since \(f^{\prime\prime}(5)>0\), \(x = 5\) is a local minimum.
Since \(f^{\prime\prime}(9)<0\), \(x = 9\) is a local maximum.

Step4: Find the function values

For \(x = 5\):
\(f(5)=-2\times5^{3}+42\times5^{2}-270\times5 + 5=-2\times125+42\times25-1350 + 5=-250+1050-1350 + 5=-545\).
For \(x = 9\):
\(f(9)=-2\times9^{3}+42\times9^{2}-270\times9 + 5=-2\times729+42\times81-2430 + 5=-1458+3402-2430 + 5=-481\).

Answer:

The function has a local minimum at \(x = 5\) with value \(-545\) and a local maximum at \(x = 9\) with value \(-481\).