Question

find the absolute minimum value on $0,\infty)$ for $f(x)=3x^{2}-18x + 6$.
select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
○ a. the absolute minimum is $\square$ at $x = \square$.
○ b. there is no absolute minimum.

Explanation:

Step1: Find the derivative of the function

The function is \(f(x)=3x^{2}-18x + 6\). Using the power rule \((x^n)^\prime=nx^{n - 1}\), the derivative \(f^\prime(x)=(3x^{2}-18x + 6)^\prime=6x-18\).

Step2: Find the critical points

Set \(f^\prime(x) = 0\), so \(6x-18=0\). Solving for \(x\):

$$
LATEXBLOCK0
$$

Step3: Check the second - derivative (to confirm it's a minimum)

The second - derivative \(f^{\prime\prime}(x)=(6x - 18)^\prime=6>0\). Since \(f^{\prime\prime}(x)>0\), \(x = 3\) is a local minimum.

Step4: Evaluate the function at the critical point and the endpoint of the interval

• Evaluate \(f(x)\) at \(x = 3\): \(f(3)=3\times(3)^{2}-18\times3 + 6=3\times9-54 + 6=27-54 + 6=-21\).
• Evaluate \(f(x)\) as \(x

ightarrow\infty\). \(\lim_{x
ightarrow\infty}f(x)=\lim_{x
ightarrow\infty}(3x^{2}-18x + 6)=\infty\) (because the leading term \(3x^{2}\) dominates as \(x
ightarrow\infty\)).

• Evaluate \(f(x)\) at \(x = 0\): \(f(0)=3\times0^{2}-18\times0 + 6=6\).

Answer:

A. The absolute minimum is \(-21\) at \(x = 3\).