Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. (Enter your answers using interval notation. If the answer cannot be expressed as an interval, enter EMPTY or ∅.)
f(x) = (5 − x^2) / x
my work:
a) f'(x)=(-x^2-5)/x^2
b) set numerator equal to zero to find critical points --> c = +/- square root of 5
c) do a number line of x-values separated by the critical points to determine increasing and decreasing --> find increasing (-square root 5, square root 5); decreasing (- infinity, -square root 5), (square root 5, infinity)
This is incorrect. Where did I go wrong?
look at the slope in the interval, it is negative, decreasin from -sqrt5 to sqrt5
then look at x=100, slope is negative again.
and at x=-100, negative again.
Look at the graph here: http://www.wolframalpha.com/input/?i=(5+%E2%88%92+x%5E2)+%2F+x
f'(x) = ( - x² - 5 ) / x² = - 1 - 5 / x² = - ( 1 + 5 / x² )
function is increasing on the interval, where f′(x) > 0
function is decreasing on the interval, where f′(x) < 0
In this case:
f'(x) = - ( 1 + 5 / x² ) is always < 0
becouse:
when x-> - ∞
then
f'(x) = - ( 1 + 5 / x² ) -> - 1
when x-> 0
then
f'(x) = - ( 1 + 5 / x² ) -> - ∞
when x-> ∞
then
f'(x) = - ( 1 + 5 / x² ) -> - 1
So f'(x) is always < 0
Your function decreasing on the interval (- ∞ ,∞ )
Thank you so much! I understand now.
Your approach is almost correct, but you made a minor mistake in determining the sign of the derivative.
Let's go through the steps again:
a) To find the derivative of f(x), you correctly computed f'(x) as (-x^2 - 5)/x^2.
b) To determine the critical points, you need to find where the numerator of the derivative equals zero. However, you made a mistake in the numerator. The numerator of the derivative should be -x^2 + 5, not -x^2 - 5. So, setting the numerator equal to zero: -x^2 + 5 = 0. Solving this equation, you get x = ±√5.
c) Now, we need to determine the intervals of increasing and decreasing. To do this, we construct a sign chart by choosing test points in each interval.
Consider intervals (-∞, -√5), (-√5, √5), and (√5, ∞).
For the interval (-∞, -√5):
Choose a test point, e.g., x = -10. Plugging this value into the derivative, f'(-10), gives (100 - 5)/100 = 95/100 = 0.95. Since f'(-10) is positive, it means the function is increasing in the interval (-∞, -√5).
For the interval (-√5, √5):
Choose a test point, e.g., x = 0. Plugging this value into the derivative, f'(0), gives (0 - 5)/0^2 = -5/0 = undefined. An undefined derivative means we cannot make any conclusions about the increasing or decreasing behavior of the function in this interval.
For the interval (√5, ∞):
Choose a test point, e.g., x = 10. Plugging this value into the derivative, f'(10), gives (100 - 5)/100 = 95/100 = 0.95. Since f'(10) is positive, it means the function is increasing in the interval (√5, ∞).
Therefore, the correct intervals are:
Increasing: (-∞, -√5) and (√5, ∞)
Decreasing: EMPTY or ∅ (since there are no intervals where the function is decreasing)
So, in summary, the function is increasing in the intervals (-∞, -√5) and (√5, ∞), and there are no intervals where it is decreasing.