Solving Mean Value Theorem: f(x)=sqrtX-2x [0,4]

[npellegrino]

1. Use the Mean Value Theorem
f(x) = sqrtX - 2x at [0,4] so a = 0, b = 4

3. So I found the derivative (which is the slope) and then set the derivative equal to the 1 because of f(b)-f(a)/b-a
f'(x) = 1/2sqrtX - 2 so 1/2sqrtX - 2 = 0, then 1/2sqrtX = 2, then my solution is x = 1/16

Just would like a heads up if this is correct, thanks in advanced.

 

[Mark44]

1. Use the Mean Value Theorem
f(x) = sqrtX - 2x at [0,4] so a = 0, b = 4

3. So I found the derivative (which is the slope) and then set the derivative equal to the 1 because of f(b)-f(a)/b-a
f'(x) = 1/2sqrtX - 2 so 1/2sqrtX - 2 = 0, then 1/2sqrtX = 2, then my solution is x = 1/16

Just would like a heads up if this is correct, thanks in advanced.

No, this isn't correct. I think you probably made a mistake in your calculation of (f(b) - f(a))/(b - a), so check your work on this calculation.

Notice also that the extra parentheses I used (and you didn't) are necessary. The correct interpretation of what you wrote would be
f(b) - \frac{f(a)}{b} - a

and I'm sure that's not what you intended.

 

[npellegrino]

well the Mean Value Theorem is

so if a = 0 and b = 4 wouldn't the solution be 1 ? then you set the derivative of the function to 1 which gave me 1/16 ?

 

[Mark44]

I think you might be calculating f'(a) and f'(b), instead of f(a) and f(b). Anyway (f(b) - f(a))/(b - a) \neq 1.

 

[cheddacheeze]

subsitute the values of a and b into the formula
find the derivative of f(x)
and after youve got f'(x) and (f(b)-f(a))/b-a
find your x

 

[npellegrino]

the derivative i got for f(x) is 1/2sqrtX - 2 then i set that equals to (f(b)-f(a))/b-a which is 1, so I'm trying to figure 1/2sqrtX - 2 = 1 and I'm having trouble figuring the fraction

 

[cheddacheeze]

how are you getting the slope = 1, i got -6/4
f(4) = 4^1/2 - 8 = -6
f(0) = 0
(-6-0)/(4-0) does not equal to 1

well using my values i get 1/(2(x^1/2)) = -6/4+2
you can probably work your way from there

 

[Mark44]

the derivative i got for f(x) is 1/2sqrtX - 2 then i set that equals to (f(b)-f(a))/b-a which is 1, so I'm trying to figure 1/2sqrtX - 2 = 1 and I'm having trouble figuring the fraction

Stop already! (f(b)-f(a))/(b-a) IS NOT EQUAL TO 1 !

Put in the actual numbers for a and b and do the calculation!

 

[npellegrino]

You are not taking the derivatives of f(b) and f(a)... the mean value theorem states that if f is continuous on the closed intervals [a,b] which in this case is [0,4] and differentiable on the open interval (a,b,) then there exists a number c in (a,b) such that

 

[npellegrino]

Stop already! (f(b)-f(a))/(b-a) IS NOT EQUAL TO 1 !

Put in the actual numbers for a and b and do the calculation!

maybe my professor has taught me wrong?

 

[cheddacheeze]

just look at the formula and substitute a and b... into the equation

 

[npellegrino]

Alright I'm having a brainfart i see what you are saying.

 

[npellegrino]

f(4) = sqrt4 - 8 = -6
f(0) = sqrt0 - 2(0) = 0

 

[npellegrino]

Alright i think i solved it... finally thanks for the help i really do appreciate it from you 2 :)f(4) = -6
f(0) = 0
which gives me -6/4

the derivative 1/2sqrtX -2 = -6/4 which will give me 1/2sqrtx = 1/2 then x = 1

 

[Mark44]

Yes.

 

[Mark44]

maybe my professor has taught me wrong?

I'm pretty sure he didn't teach you that (2 - 8)/(4 - 0) = 1.

 

[cheddacheeze]

I'm pretty sure he didn't teach you that (2 - 8)/(4 - 0) = 1.

not so sure about that, sometimes the things they teach you are so obstract that they themselves get confused

 

[Mark44]

No question that an instructor can occasionally get confused with some abstract calculation, but (2 - 8)/(4 - 0) in no way can be considered as abstract.

 

[cheddacheeze]

indeed it is not abstract, the mean value theorem is very elementary since its dealing with slopes and derivatives so its easy to understand