Evaluating Trigonometric Expressions

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Discussion Overview

The discussion revolves around evaluating the derivative of the function h(x) = tan x over the interval [π/4, 1]. Participants explore the nature of the problem, whether it involves calculus concepts such as the Mean Value Theorem or the Definition of the Derivative, and how to approach the evaluation of the function over the specified range.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants seek hints on how to evaluate dh/dx for h(x) = tan x on the interval [π/4, 1].
  • There is uncertainty about whether the problem requires the Definition of the Derivative or if it is a Mean Value problem.
  • One participant proposes using the formula for the average rate of change: $\dfrac{\Delta h}{\Delta x} = \dfrac{h(1) - h\left(\frac{\pi}{4}\right)}{1 - \frac{\pi}{4}}$.
  • Another participant asserts that no derivative is required and clarifies that the question is from a precalculus textbook, suggesting that calculus is not necessary.
  • Some participants agree that the problem can be viewed as a Mean Value problem, while others question if it is more accurately described as a rate of change problem.
  • There is a discussion about the terminology, with one participant stating that the Average Rate of Change is not the same as the Mean Value, referencing the Mean Value Theorem.

Areas of Agreement / Disagreement

Participants express differing views on whether the problem is a Mean Value problem or simply an average rate of change problem. There is no consensus on the necessity of calculus for solving the problem, as some argue it is not required while others suggest it may be relevant.

Contextual Notes

Participants mention a textbook context, which may influence the interpretation of the problem. The discussion reflects varying levels of familiarity with calculus concepts and terminology.

Who May Find This Useful

Students studying precalculus or introductory calculus concepts, particularly those interested in the evaluation of trigonometric functions and the Mean Value Theorem.

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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