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

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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