Question

find a) any critical values and b) any relative extrema. f(x)=x^2 - 6x + 3 a) select the correct choice below and, if necessary, fill in the answer box within your choice. a. the critical value(s) of the function is/are (use a comma to separate answers as needed.) b. the function has no critical values.

Explanation:

Step1: Find the derivative of the function

$f(x)=x^{2}-6x + 3$, using the power - rule $(x^n)'=nx^{n - 1}$, we have $f'(x)=2x-6$.

Step2: Set the derivative equal to zero to find critical values

Set $f'(x)=0$, so $2x - 6=0$. Solving for $x$:
Add 6 to both sides: $2x=6$. Then divide by 2, we get $x = 3$.

Step3: Use the second - derivative test to find relative extrema

Find the second - derivative $f''(x)$. Since $f'(x)=2x - 6$, then $f''(x)=2>0$.
When $f''(x)>0$ at a critical value $x = c$, the function has a relative minimum at $x = c$.
Substitute $x = 3$ into the original function $f(x)$: $f(3)=3^{2}-6\times3 + 3=9-18 + 3=-6$.

Answer:

a) A. The critical value(s) of the function is/are $3$.
b) The relative minimum is $-6$ at $x = 3$.