Evaluating Trigonometric Expressions

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SUMMARY

The discussion centers on evaluating the derivative of the function h(x) = tan x over the interval [π/4, 1]. Participants clarify that this problem does not require calculus, as it pertains to the Mean Value Theorem and the Average Rate of Change. The formula used for evaluation is Δh/Δx = (tan(1) - tan(π/4)) / (1 - π/4). The conversation highlights the distinction between Average Rate of Change and Mean Value, emphasizing the importance of understanding these concepts in precalculus.

PREREQUISITES
  • Understanding of the Mean Value Theorem
  • Familiarity with Average Rate of Change
  • Basic knowledge of trigonometric functions, specifically tangent
  • Ability to interpret and manipulate expressions in calculus
NEXT STEPS
  • Study the Mean Value Theorem in detail
  • Learn how to calculate Average Rate of Change for various functions
  • Explore trigonometric derivatives and their applications
  • Practice problems from precalculus textbooks, particularly Section 6.3
USEFUL FOR

Students in precalculus, educators teaching trigonometric functions, and anyone seeking to deepen their understanding of the Mean Value Theorem and its applications in calculus.

mathdad
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Given h(x) = tan x, evaluate dh/dx on [pi/4, 1].

Note: d = delta

I need one or two hints. I can then try on my own.
 
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RTCNTC said:
Given h(x) = tan x, evaluate dh/dx on [pi/4, 1].

Note: d = delta

I need one or two hints. I can then try on my own.
You'll have to do better than that. Are we required to use the Definition of the Derivative? Is this a Mean Value problem? How does one "evaluate" on a range?
 
RTCNTC said:
Given h(x) = tan x, evaluate dh/dx on [pi/4, 1].

Note: d = delta

I need one or two hints. I can then try on my own.

$\dfrac{\Delta h}{\Delta x} = \dfrac{h(1) - h\left(\frac{\pi}{4}\right)}{1 - \frac{\pi}{4}}$

Note: learn some Latex ...
 
tkhunny said:
You'll have to do better than that. Are we required to use the Definition of the Derivative? Is this a Mean Value problem? How does one "evaluate" on a range?

Skeeter got it. No derivative required. This question comes from my precalculus textbook by David Cohen. No calculus needed here. I was just not sure how to start the calculation. Like I said, Skeeter got it.
 
Ok.

The set up is:

[tan 1 - tan (π/4)]/(1 - π/4)

I will do it tonight after work.
 
So it is a Mean Value problem. Fair enough.
 
tkhunny said:
So it is a Mean Value problem. Fair enough.

Is it more a rate of change problem?
 
RTCNTC said:
Is it more a rate of change problem?

Right. An Average Rate of Change, aka Mean Value. Anyway it is done,
 
It is done as far as this question is concerned. There are about 5 or 6 similar questions in Section 6.3 in my book that I can now do thanks to this site.
 
  • #10
tkhunny said:
Right. An Average Rate of Change, aka Mean Value. Anyway it is done,

An Average Rate of Change is NOT called a Mean Value.

The Mean Value Theorem states that the Average Rate of Change is equal to the gradient of the function at some point within the interval. It's poorly named too, as it is most likely not the "mean value" which gives that same gradient.
 
  • #11
Good information.
 

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