00 points possible answered: 3/10 question 4 the function ( f(x)=2 x^{3}-30 x^{2}+144 x + 10 ) has one local…

Explanation:

Step1: Find the first derivative

The derivative of (f(x)=2x^{3}-30x^{2}+144x + 10) is (f^{\prime}(x)=6x^{2}-60x + 144).

Step2: Set the first derivative equal to zero

Set (f^{\prime}(x)=0), so (6x^{2}-60x + 144 = 0). Divide through by (6) to get (x^{2}-10x + 24=0). Factor the quadratic: ((x - 4)(x - 6)=0). The critical points are (x = 4) and (x = 6).

Step3: Find the second derivative

The second derivative (f^{\prime\prime}(x)=12x-60).

Step4: Use the second - derivative test

• For (x = 4): (f^{\prime\prime}(4)=12\times4-60=48 - 60=-12<0). So (x = 4) is a local maximum.
• For (x = 6): (f^{\prime\prime}(6)=12\times6-60=72 - 60 = 12>0). So (x = 6) is a local minimum.

Step5: Find the function values

• When (x = 6): (f(6)=2\times6^{3}-30\times6^{2}+144\times6 + 10=2\times216-30\times36 + 864+10=432-1080 + 864+10=226).
• When (x = 4): (f(4)=2\times4^{3}-30\times4^{2}+144\times4 + 10=2\times64-30\times16 + 576+10=128-480 + 576+10=234).

Answer:

The function has a local minimum at (x = 6) with function value (226) and a local maximum at (x = 4) with function value (234).