How do you sketch the graph #y=x+1/x# using the first and second derivatives?

Question

Isabella Ackerman · Accepted Answer

See explanation that includes graph that is a rectangular hyperbola. with asymptotes x = 0 and y = 1.

#y'=1-1/x^2=0, when x=+-1.# #y''=2/x^3 =-2#, at x =# -1 # and #2# at x = 1# Interestingly, no higher derivative is 0 at #x=+-1#.. So, there are no maxima and minima for y. \ As # x to +- oo, y to+-oo#. As #y to 1, x to+-oo#. So, x =0 and y = 1 are the asymptotes. graph{xy-x-1=0 [-10, 10, -5, 5]}

Charles Anctil · Answer

undefined To sketch the graph of $ y = x + \frac{1}{x} $ using the first and second derivatives, follow these steps: 1. Find the first derivative of $ y $ with respect to $ x $. 2. Find critical points by setting the first derivative equal to zero and solving for $ x $. 3. Use the first derivative test to determine the increasing and decreasing intervals. 4. Find the second derivative of $ y $ with respect to $ x $. 5. Use the second derivative test to determine concavity and inflection points. 6. Sketch the graph based on the information obtained from steps 3 and 5. Let's start with finding the first derivative: $$ \frac{dy}{dx} = 1 - \frac{1}{x^2} $$ Setting this derivative equal to zero: $$ 1 - \frac{1}{x^2} = 0 $$ Solving for $ x $: $$ x^2 - 1 = 0 \implies x = \pm 1 $$ These are the critical points. Now, we use the first derivative test to determine increasing and decreasing intervals: - For $ x < -1 $ and $ x > 1 $, $ \frac{dy}{dx} > 0 $, so the function is increasing. - For $ -1 < x < 1 $, $ \frac{dy}{dx} < 0 $, so the function is decreasing. Next, we find the second derivative: $$ \frac{d^2y}{dx^2} = \frac{2}{x^3} $$ Now, we check the concavity and inflection points: - The second derivative is positive for $ x < 0 $ and negative for $ x > 0 $, indicating that the function is concave up for $ x < 0 $ and concave down for $ x > 0 $. - There are no inflection points. Finally, we sketch the graph using the information obtained: - For $ x < -1 $, the function is increasing and concave up. - For $ -1 < x < 0 $, the function is decreasing and concave up. - For $ 0 < x < 1 $, the function is decreasing and concave down. - For $ x > 1 $, the function is increasing and concave down. Combine this information to sketch the graph of $ y = x + \frac{1}{x} $.