How do you find intervals where the graph of #f(x) = x + 1/x# is concave up and concave down?

Answer 1
When you take the first derivative (#d/(dx)#) and set it equal to 0, you can find where the slope is 0, and that lets you determine at what points on the graph that there are concavities. Taking the second derivative (#d^2/(dx^2)#) tells you whether it is concave up (positive) or down (negative).
#f(x) = 1 + x^-1#
#d/(dx)[f(x)] = 1 + (-1x^-2) = 1 - 1/(x^2) = 0#
#1 = 1/(x^2) => x^2 = 1 => x = 1, x = -1#
Plug it back into #f(x)# to get:
#f(1) = 2# #f(-1) = -2#
Thus, your two coordinates for the extrema are #(-1,-2)# and #(1,2)#. Now, let's take the second derivative to find their concavity directions.
#d^2/(dx^2)[f(x)] = -(-2x^(-3)) = 2/(x^3)#
We don't have to set it to 0 here. If we plug in #1# and #-1# from the earlier first derivative, we get:
#(d^2)/(dx^2)[f(1)] = 2# #(d^2)/(dx^2)[f(-1)] = -2#
Since the first one is positive, the coordinate #(1, 2)# is where the concave-up part of the curve changes direction, and since the second one is negative, the coordinate #(-1,-2)# is where the concave-down part of the curve changes direction.
Naturally, an asymptote is #x = 0#. Another asymptote is #x = y#. Let's try it:
#y = y + 1/y => 0 = 1/y => y ne 1/0 = "undefined"#
Anyways, since there's a vertical asymptote at #x = 0#, the domain is (interval notation):
#(-oo, 0) uu (0, oo)#
So, the graph is therefore concave down at #(-oo, 0)#, and concave up at #(0, oo)#.
Overall, you know what #y = x# and #y = 1/x# look like; adding them together has #y = x# transform #y = 1/x# by giving it a #y = x# asymptote, like so: graph{x+1/x [-10, 10, -5, 5]}
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Answer 2
To find intervals where the graph of \( f(x) = x + \frac{1}{x} \) is concave up and concave down, we need to analyze the second derivative of the function. First, find the first derivative of \( f(x) \) using the sum and power rules: \[ f'(x) = 1 - \frac{1}{x^2} \] Now, find the second derivative by differentiating \( f'(x) \): \[ f''(x) = \frac{2}{x^3} \] To determine where the graph is concave up or down, we need to find where \( f''(x) > 0 \) (concave up) and where \( f''(x) < 0 \) (concave down). Since \( f''(x) = \frac{2}{x^3} \), the second derivative is positive when \( x > 0 \) and negative when \( x < 0 \). Therefore, the graph of \( f(x) = x + \frac{1}{x} \) is concave up on the interval \( x > 0 \) and concave down on the interval \( x < 0 \).
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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