Question

section 2.8: optimization (homework)
score: 0/120 answered: 0/12
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question 1
0/10 pts 4 99 details
the function $f(x)=-2x^{3}+33x^{2}-60x + 8$ has one local minimum and one local maximum.
this function has a local minimum at $x =$
with value
and a local maximum at $x =$
with value
question help: video

Explanation:

Step1: Find the derivative

Differentiate $f(x)=- 2x^{3}+33x^{2}-60x + 8$ using the power - rule. The derivative $f'(x)=-6x^{2}+66x - 60$.

Step2: Set the derivative equal to zero

Solve $-6x^{2}+66x - 60 = 0$. Divide through by $-6$ to get $x^{2}-11x + 10=0$. Factor the quadratic equation: $(x - 1)(x - 10)=0$. So the critical points are $x = 1$ and $x = 10$.

Step3: Use the second - derivative test

Find the second - derivative $f''(x)=-12x + 66$.
When $x = 1$, $f''(1)=-12\times1+66 = 54>0$. So $f(x)$ has a local minimum at $x = 1$.
When $x = 10$, $f''(10)=-12\times10+66=-54<0$. So $f(x)$ has a local maximum at $x = 10$.

Step4: Find the function values

For the local minimum at $x = 1$: $f(1)=-2\times1^{3}+33\times1^{2}-60\times1 + 8=-2 + 33-60 + 8=-21$.
For the local maximum at $x = 10$: $f(10)=-2\times10^{3}+33\times10^{2}-60\times10 + 8=-2000+3300-600 + 8=708$.

Answer:

Local minimum at $x = 1$ with value $-21$; local maximum at $x = 10$ with value $708$