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Use the second derivative to find concavity.
It's concave down where f'' is negative, ie f'' < 0
Concave up where f'' is positive, f'' > 0
Find where f'' = 0 firstly.
Did you find f''? once you do, you can find where it's concave up and down.
Just from looking at the graph, I'd guess concave down between about -inf and -1, and +1 and +infinity
Concave up between about -1 and 1.
Once you get the x values where f'' = 0, pick points to the left and right of those x values, and check if f'' is positive (concave down) or negative (concave up) to help find the intervals.
@Abarnett probably easier to just link to one than hand draw a graph :P
f'': https://www.google.com/search?q=10(1-3x%5E2)%2F(x%5E2+%2B1)%5E3&aq=f&oq=10(1-3x%5E2)%2F(x%5E2+%2B1)%5E3&aqs=chrome.0.57j60l3j62l2.174j0&sourceid=chrome&ie=UTF-8
Oh and @Jgeurts f'' can also be used to find if a critical point (where f' = 0) is a max or min - at a maximum, f'' is negative. At a minimum, f'' is positive.
But that's part of the concavity anyway, since maximums are concave down, minimums are concave up.
If f'' is zero at a point where f' = 0, then it's an inflection point.
@Abarnett@agent0smith So I have
Global max (3,1/2), Global Min (0,-4),
Decreasing on the interval [-2,0], Increasing on the interval [0,3],
local max is f(3), local min is f(0),
concave down at [-inf,-sqrt1/3]U[sqrt1/3, inf],
concave up at [-sqrt1/3, sqrt1/3],
inflection points occur at +-sqrt1/3
lol that interval [-2,3] makes a big difference... as the original function has no maximum, it just has an asymptote.
https://www.google.com/search?q=(x%5E2+-+4)%2F(x%5E2+%2B1)&aq=f&oq=(x%5E2+-+4)%2F(x%5E2+%2B1)&aqs=chrome.0.57j60l3j62l2.269j0&sourceid=chrome&ie=UTF-8
That's the original function... all your values look good.
Except:
concave down at [-inf,-sqrt1/3]U[sqrt1/3, inf],
Something seems off here, with an interval [-2,3] ;)
If you can learn to kinda read the graph, you can check your values. The max/mins should be easy to find (turning points), same with where the graph is increasing or decreasing (the slope f' is positive or negative).
Points of inflection are more difficult to see... you kinda have to gauge where it changes from concave up to down which can be hard to see, but still possible to estimate where it is.
Correct, since f' will never equal 0 except at the minimum.
\[\large f'(x) = 10x/(x^2 +1)^2\] - note that this can be only zero if x=0 (the denominator can never make the f' equal zero) - ie there's no other turning points/maximums, only a minimum.
Ok, i feel so informed, haha.
Hold up, last question! What about asymptotes? Horizontal/Vertical? Obviously theres one around y=1 but how do i prove it?
To find asymptotes... you just have to look at the original function, there's no calculus involved.
\[\large f(x) = (x^2 - 4)/(x^2 +1)\]
What happens when x gets really huge (positive or negative infinity? The -4 and +1 become tiny compared to the huge x^2, so it becomes
\[\Large \frac{ x^2 }{ x^2 }\] which is equal to ...?
Correct! :) that's for horizontal asymptotes. So y= 1 is a horizontal asymptote.
To find vertical asymptotes, find values of x that cause the denominator to be zero
\[\Large f(x) = \frac{ (x^2 - 4)}{(x^2 +1)}\]are there any x values that make the denominator zero? x^2 is always positive, and you're adding 1...
What about instead for this function?
\[\Large f(x) = \frac{ (x^2 - 4)}{(x +1)}\]
For x^3/x^2, that simplifies to just x - it's actually a slant asymptote (neither horizontal or vertical) - there is no horizontal. BUT x can't equal zero, so it has a vertical at x=0 (or just a 'hole' at x=0, not really an asymptote in this case), since the denominator is zero.
Yep. With a hole at x=0.
So for vertical - look for values of x that give a denominator zero
For horizontal - look for how the function behaves when x is approaching +infinity or -infinity.