Question

find a) any critical values and b) any relative extrema.
f(x)=x² - 2x + 5
a) select the correct choice below and, if necessary, fill in the answer box within your choice.
oa. the critical value(s) of the function is/are
(use a comma to separate answers as needed )
ob. the function has no critical values.
b) select the correct choice below and, if necessary, fill in the answer box(es) within your choice.
oa. the relative maximum point(s) is/are and there are no relative minimum points
(simplify your answer. type an ordered pair, using integers or fractions. use a comma to separate
answers as needed )
ob. 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 )
oc. the relative minimum 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.)
od. there are no relative minimum points and there are no relative maximum points.

Explanation:

Step1: Find the derivative

The derivative of \(f(x)=x^{2}-2x + 5\) using the power rule \((x^n)^\prime=nx^{n - 1}\) is \(f^\prime(x)=2x-2\).

Step2: Find critical values

Set \(f^\prime(x) = 0\). So, \(2x-2=0\). Add \(2\) to both sides: \(2x=2\). Divide both sides by \(2\): \(x = 1\).

Step3: Use the second - derivative test

Find the second derivative. \(f^{\prime\prime}(x)=(2x - 2)^\prime=2\). Since \(f^{\prime\prime}(1)=2>0\), the function has a relative minimum at \(x = 1\).
Substitute \(x = 1\) into \(f(x)\): \(f(1)=1^{2}-2\times1 + 5=1-2 + 5=4\).

Answer:

a) A. The critical value(s) of the function is/are \(1\)
b) C. The relative minimum point(s) is/are \((1,4)\) and there are no relative maximum points