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

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The discussion focuses on applying the Mean Value Theorem to the function f(x) = √x over the interval [0, 4]. The calculation shows that the secant line's slope between the endpoints is 1/2, and the derivative f'(x) is also 1/2 at x = 1, confirming that c = 1 satisfies the theorem. The tangent line at (1, 1) is parallel to the secant line, as both have the same slope. Participants discuss the clarity and elegance of the solution method, with some uncertainty about the presentation of the steps. Overall, the conclusion is that the secant and tangent lines are indeed parallel at the specified point.
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.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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