Question

find any relative extrema of the function. list each extremum along with the x - value at which it occurs. identify intervals over which the function is increasing and over which it is decreasing. then sketch a graph of the function.

$f(x)=1 - 3x+x^{2}$

describe any relative extrema. select the correct choice below and, if necessary, fill in the answer boxes in within your choice.

a. the relative maximum point(s) is/are and there are no relative maximum points.
(simplify your answer. type an ordered pair, using integers or fractions. use a comma to separate answers as needed.)

b. the relative minimum point(s) is/are and the relative maximum point(s) is/are
(simplify your answers. type ordered pairs, using integers or fractions. use a comma to separate answers as needed.)

c. the relative maximum point(s) is/are and there are no relative minimum points.
(simplify your answer. type as ordered pair, using integers or fractions. use a comma to separate answers as needed.)

d. there are no relative minimum points and there are no relative maximum points.

Explanation:

Step1: Find the derivative

Given $f(x)=1 - 3x+x^{2}$, the derivative $f'(x)=\frac{d}{dx}(1 - 3x+x^{2})$. Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, we have $f'(x)=-3 + 2x$.

Step2: Find the critical points

Set $f'(x)=0$. So, $-3 + 2x=0$. Solving for $x$, we get $2x=3$, then $x=\frac{3}{2}$.

Step3: Determine the second - derivative

Find the second - derivative $f''(x)=\frac{d}{dx}(-3 + 2x)=2$. Since $f''(\frac{3}{2}) = 2>0$, the function has a relative minimum at $x=\frac{3}{2}$.

Step4: Find the value of the relative minimum

Substitute $x = \frac{3}{2}$ into the original function $f(x)$: $f(\frac{3}{2})=1-3\times\frac{3}{2}+(\frac{3}{2})^{2}=1-\frac{9}{2}+\frac{9}{4}=\frac{4 - 18+9}{4}=-\frac{5}{4}$.

Step5: Determine intervals of increase and decrease

Choose a test point in the interval $(-\infty,\frac{3}{2})$, say $x = 0$. Then $f'(0)=-3<0$, so the function is decreasing on the interval $(-\infty,\frac{3}{2})$. Choose a test point in the interval $(\frac{3}{2},\infty)$, say $x = 2$. Then $f'(2)=1>0$, so the function is increasing on the interval $(\frac{3}{2},\infty)$.

Answer:

B. The relative minimum point(s) is $(\frac{3}{2},-\frac{5}{4})$ and there are no relative maximum point(s).