find the absolute minimum value on $0,\\infty)$ for $f(x)=3x^{2}-18x + 6$.\nselect the correct choice below…

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

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

Answer

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): [ \begin{align*} 6x&=18\ x&=3 \end{align*} ]

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