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

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SUMMARY

The discussion focuses on applying the Mean Value Theorem (MVT) to the function \( f(x) = \sqrt{x} \) over the interval \([0, 4]\). The calculated value of \( c \) that satisfies the MVT is \( c = 1 \), where \( f(c) = 1 \). The slope of the secant line is determined to be \( m = \frac{1}{2} \), and the derivative \( f'(x) = \frac{1}{2\sqrt{x}} \) confirms that \( f'(1) = \frac{1}{2} \). The secant line and the tangent line at this point are indeed parallel.

PREREQUISITES
  • Understanding of the Mean Value Theorem (MVT)
  • Basic knowledge of derivatives and their calculations
  • Graphing functions and lines in a Cartesian coordinate system
  • Familiarity with the function \( f(x) = \sqrt{x} \)
NEXT STEPS
  • Study the application of the Mean Value Theorem in different functions
  • Learn how to graph secant and tangent lines for various functions
  • Explore the implications of parallel lines in calculus
  • Investigate the conditions under which the Mean Value Theorem holds
USEFUL FOR

Students of calculus, mathematics educators, and anyone interested in understanding the Mean Value Theorem and its graphical interpretations.

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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