f(x)=x^{2}-2x - 63. (a) find the interval(s) where f(x) is increasing. (b) find the interval(s) where f(x)…

f(x)=x^{2}-2x - 63. (a) find the interval(s) where f(x) is increasing. (b) find the interval(s) where f(x) is decreasing. (c) find the x - value(s) of all relative maxima of f(x). if there are none, enter none. if there are multiple relative maxima, separate the values with commas. x = (d) find the x - value(s) of all relative minima of f(x). if there are none, enter none. if there are multiple relative minima, separate the values with commas.
Answer
Explanation:
Step1: Find the derivative
The derivative of $f(x)=x^{2}-2x - 63$ using the power - rule $(x^n)'=nx^{n - 1}$ is $f'(x)=2x-2$.
Step2: Find the critical points
Set $f'(x) = 0$. So, $2x-2=0$. Solving for $x$ gives $2x=2$, then $x = 1$.
Step3: Test the intervals
Choose a test point in the interval $(-\infty,1)$, say $x = 0$. Then $f'(0)=2(0)-2=-2<0$. So, $f(x)$ is decreasing on $(-\infty,1)$. Choose a test point in the interval $(1,\infty)$, say $x = 2$. Then $f'(2)=2(2)-2 = 2>0$. So, $f(x)$ is increasing on $(1,\infty)$. Since the function changes from decreasing to increasing at $x = 1$, there is a relative minimum at $x = 1$ and no relative maximum.
Answer:
(a) $(1,\infty)$ (b) $(-\infty,1)$ (c) NONE (d) $1$