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

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).