Mean Value Theorem: Show that |tanA−tanB|≥|A−B| for all A and B

[Win_odd Dhamnekar]

Hello,
I want to show that $\displaystyle |tanA-tanB| \geq |A-B|$ for all A and B ( in radians )in(-π/2,π/2).Can the inequality be extended to all A and B?
Solution:- Here how can I use Mean Value Theorem?

 

[Dr.Peterson]

Hello,
I want to show that $\displaystyle |tanA-tanB| \geq |A-B|$ for all A and B ( in radians )in(-π/2,π/2).Can the inequality be extended to all A and B?
Solution:- Here how can I use Mean Value Theorem?

I take it you have reason to think the MVT might be relevant, such as that you just learned about it. You're right; it is.

What would the MVT say for this particular function, applied to an interval [A,B]?

What has to be true of the derivative of the tangent in order for this to imply the desired inequality?

Let us know what you are thinking, so we can help you further.

 

[Steven G]

Hello,
I want to show that $\displaystyle |tanA-tanB| \geq |A-B|$ for all A and B ( in radians )in(-π/2,π/2).Can the inequality be extended to all A and B?

Don't be so lazy, please pick some values for A and B and see if the inequality holds each time. Of course, if they do then you only have a conjecture and will have to prove. If you find values for A and B that do not verify the inequality, then no it does not hold for all A and B.
Maybe you can even use some symmetry to conclude if the conjecture is true.
Please post back even if just to ask for help.

 

[Win_odd Dhamnekar]

Don't be so lazy, please pick some values for A and B and see if the inequality holds each time. Of course, if they do then you only have a conjecture and will have to prove. If you find values for A and B that do not verify the inequality, then no it does not hold for all A and B.
Maybe you can even use some symmetry to conclude if the conjecture is true.
Please post back even if just to ask for help.

Hello,
Let us see is it true for interval [-1,1]. |-1-1|=2 |tan-1-tan1|=|-1.55740772-1.55740772|=3.11481544. Hence we can say $\displaystyle |tanA-tanB|\geq |A-B|$ is true.

 

[Win_odd Dhamnekar]

I take it you have reason to think the MVT might be relevant, such as that you just learned about it. You're right; it is.

What would the MVT say for this particular function, applied to an interval [A,B]?

What has to be true of the derivative of the tangent in order for this to imply the desired inequality?

Let us know what you are thinking, so we can help you further.

hello,
$\displaystyle f'(c)=\frac {tanB-tanA}{B-A}=\frac{3.114815}{2}$ Derivative of $\displaystyle Tan(x)=sec^2(x)$ x=0.6413199 which lies in the interval [-1,1]. So i have shown $\displaystyle |tanA-tanB|\geq|A-B|$

 

[pka]

Hello,
I want to show that $\displaystyle |tanA-tanB| \geq |A-B|$ for all A and B ( in radians )in(-π/2,π/2).Can the inequality be extended to all A and B?
Solution:- Here how can I use Mean Value Theorem?

Have you carefully reviewed the mean value theorem?
Have you carefully reviewed the graph of the $\displaystyle \tan(x)$

If you have then you understand that for $\displaystyle [A,B] \subset \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$ the statement holds, but not in general.

 

[Dr.Peterson]

Hello,
Let us see is it true for interval [-1,1]. |-1-1|=2 |tan-1-tan1|=|-1.55740772-1.55740772|=3.11481544. Hence we can say $\displaystyle |tanA-tanB|\geq |A-B|$ is true.

One example doesn't prove it is always true. You can guess that, since you were told to try an example, you were expected to discover a counterexample -- one that shows it is not always true.

But it would be better to think about when the MVT applies, and why it doesn't apply here to larger intervals.

 

[Dr.Peterson]

hello,
$\displaystyle f'(c)=\frac {tanB-tanA}{B-A}=\frac{3.114815}{2}$ Derivative of $\displaystyle Tan(x)=sec^2(x)$ x=0.6413199 which lies in the interval [-1,1]. So i have shown $\displaystyle |tanA-tanB|\geq|A-B|$

I don't understand. How can you be getting a specific value with no specific values for A or B or c?

You are supposed to show that $\displaystyle |\tan A-\tan B| \geq |A-B|$ for any A and B in the given interval. The MVT says that, with $\displaystyle f(x) = tan(x)$, for some c between A and B, $\displaystyle f'(c)=\frac {\tan B-\tan A}{B-A}$, or equivalently $\displaystyle f'(c)=\frac {\tan A-\tan B}{A-B}$.

What values can $\displaystyle f'(c)$ take?

What does that tell you about $\displaystyle |\tan A-\tan B|$?