Question

section 2.8: optimization (homework) answered: 7/12 score: 79.5/120 question 9 the function (f(x)=3 + 3x+27x^{-1}) has one local minimum and one local maximum. this function has a local maximum at (x =) with value and a local minimum at (x =) with value question help: video submit question

Explanation:

Step1: Find the derivative

The derivative of $f(x)=3 + 3x+27x^{-1}$ using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$ is $f^\prime(x)=3-27x^{-2}=3-\frac{27}{x^{2}}$.

Step2: Set the derivative equal to zero

Set $f^\prime(x) = 0$, so $3-\frac{27}{x^{2}}=0$. Then $\frac{27}{x^{2}}=3$, which gives $x^{2}=9$, and $x=\pm3$.

Step3: Find the second - derivative

The second - derivative $f^{\prime\prime}(x)=\frac{54}{x^{3}}$.

Step4: Use the second - derivative test

When $x = 3$, $f^{\prime\prime}(3)=\frac{54}{3^{3}}=\frac{54}{27}=2>0$. So $f(x)$ has a local minimum at $x = 3$. Substitute $x = 3$ into $f(x)$: $f(3)=3 + 3\times3+27\times3^{-1}=3 + 9 + 9=21$.
When $x=-3$, $f^{\prime\prime}(-3)=\frac{54}{(-3)^{3}}=\frac{54}{-27}=-2<0$. So $f(x)$ has a local maximum at $x=-3$. Substitute $x=-3$ into $f(x)$: $f(-3)=3+3\times(-3)+27\times(-3)^{-1}=3-9 - 9=-15$.

Answer:

The local maximum is at $x=-3$ with value $-15$; the local minimum is at $x = 3$ with value $21$.