How do you determine if a function is concave up/ concave down if #tanx+2x #on #(-pi/2, pi/2)#?

Answer 1

Investigate the sign of the second derivative.

Let: #y = tanx + 2x# on #(-pi/2, pi/2)#
#y' = sec^2x +2# (For finding #y''#, remember that #sec^2x = (secx)^2# so we'll use the chain rule)
#y'' = 2secx secx tanx # (The derivative of #2# is #0#.)
So #y'' = 2sec^2x tanx#.
#y''# is never undefined on #(-pi/2, pi/2)#, so its only chance to change sign is at its zero(s)..
#abs(sec x) >= 1#, so #sec^2x > 1# So the only zero of #y''# is where #tanx = 0#, which is at #x = 0#
For #x# in #(-pi/2, 0)#, #tanx # is negative, so #y''# is negative and the graph of the function is concave down.
For #x# in #(0, pi/2)#, #tanx # is positive, so #y''# is positive and the graph of the function is concave up.
The function is concave down on #(-pi/2, 0)#, and it is
concave up on #(0, pi/2)#.
The point #(0,0)# is an inflection point.
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Answer 2

To determine the concavity of the function ( f(x) = \tan(x) + 2x ) on the interval ( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) ), you need to find the second derivative of the function and then analyze its sign within the given interval.

First, find the first derivative ( f'(x) ) of ( f(x) ), then find the second derivative ( f''(x) ).

[ f'(x) = \sec^2(x) + 2 ]

[ f''(x) = \frac{d}{dx}(\sec^2(x) + 2) ]

[ = \frac{d}{dx}(\sec^2(x)) + \frac{d}{dx}(2) ]

[ = 2\sec^2(x)\tan(x) ]

Now, you need to determine the sign of ( f''(x) ) within the interval ( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) ). You can observe that ( \sec^2(x) ) is always positive within this interval. And ( \tan(x) ) is positive in the interval ( \left(-\frac{\pi}{2}, 0\right) ) and negative in the interval ( \left(0, \frac{\pi}{2}\right) ). Since ( \sec^2(x) ) is always positive, the sign of ( f''(x) ) depends solely on ( \tan(x) ).

In the interval ( \left(-\frac{\pi}{2}, 0\right) ), ( f''(x) ) is positive, indicating concavity up.

In the interval ( \left(0, \frac{\pi}{2}\right) ), ( f''(x) ) is negative, indicating concavity down.

Therefore, the function ( f(x) = \tan(x) + 2x ) is concave up on ( \left(-\frac{\pi}{2}, 0\right) ) and concave down on ( \left(0, \frac{\pi}{2}\right) ).

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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.

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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