3.2.15 mvt - Mean value theorem: graphing the secant and tangent lines

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Discussion Overview

The discussion revolves around applying the Mean Value Theorem to the function \( f(x) = \sqrt{x} \) over the interval \([0, 4]\). Participants explore the calculation of the secant and tangent lines, and whether these lines are parallel.

Discussion Character

  • Mathematical reasoning
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • The original poster (OP) calculates the slope of the secant line as \( m = \frac{1}{2} \) and finds \( c = 1 \) where the tangent line's slope matches this value.
  • Some participants express uncertainty about whether the method used by the OP is the most appropriate or elegant, questioning if the correctness of approaches varies over time.
  • Another participant interprets the OP's comment about the "de jour" way as a concern for mathematical justification of the steps taken.

Areas of Agreement / Disagreement

There is no clear consensus on the appropriateness of the method used by the OP, as some participants question the clarity and justification of the steps taken, while others affirm the approach without challenge.

Contextual Notes

The discussion includes expressions of uncertainty regarding the elegance and justification of mathematical steps, but does not resolve these concerns.

Who May Find This Useful

Students and educators interested in the Mean Value Theorem, mathematical justification of problem-solving steps, and the nuances of mathematical communication may find this discussion relevant.

karush
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$\tiny{3.2.15}$
Find the number c that satisfies the conclusion of the Mean Value Theorem on the given interval. Graph the function the secant line through the endpoints, and the tangent line at $(c,f(c))$.
$f(x)=\sqrt{x} \quad [0,4]$
Are the secant line and the tangent line parallel?
$\dfrac{{f(b)-f(a)}}{{b-a}}$
then
$f(0)=0 \quad f(4)=2 \quad m=\dfrac{1}{2}$
then
$f'(x)=\dfrac{1}{2\sqrt{x}}
=\dfrac{1}{2}\quad\therefore \quad f'(1)=\dfrac{1}{2}$
then
$(c,f(c))=(1,1)$

ok not sure if this is the de jour way but...
 
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karush said:
$\tiny{3.2.15}$
Find the number c that satisfies the conclusion of the Mean Value Theorem on the given interval. Graph the function the secant line through the endpoints, and the tangent line at $(c,f(c))$.
$f(x)=\sqrt{x} \quad [0,4]$
Are the secant line and the tangent line parallel?
$\dfrac{{f(b)-f(a)}}{{b-a}}$
then
$f(0)=0 \quad f(4)=2 \quad m=\dfrac{1}{2}$
then
$f'(x)=\dfrac{1}{2\sqrt{x}}
=\dfrac{1}{2}\quad\therefore \quad f'(1)=\dfrac{1}{2}$
then
$(c,f(c))=(1,1)$

ok not sure if this is the de jour way but...

It's fine, well done.
 
"De jour" way? Do you think the correct way to do a problem changes from day to day?
 
HallsofIvy said:
"De jour" way? Do you think the correct way to do a problem changes from day to day?

I think the OP just means they are not sure if this is considered the most concise or elegant way, or if there are any steps that are not mathematically justified.
 

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