What, if anything, does the Mean Value Theorem guarantee for the given function on the given interval?
f(x)=x^2-2x+5 on [1,4]

Question

What, if anything, does the Mean Value Theorem guarantee for the given function on the given interval?
f(x)=x^2-2x+5 on [1,4]

jiteshmeghwal9 · Answer

Find f(1)=f(4) if they r equal then there is atleast one number c in the interval [1,4] for which f(c)=0

mathmale · Answer

I'd suggest you review your textbook's presentation of the M. V. T.  This Theorem involves an equation, but also two conditions that must be satisfied for the M. V. T.'s conclusion to be valid.  

Focusing on the equation:  It states that the slope of the line connecting the end points of your graph on the given interval equals the derivative of your function evaluated at some number c.  Your job is to find the vaue of c.

$$\frac{ f(b)-f(a) }{ b-a }=f '(c)$$

mathmale · Answer

Condition 1 for the MVT:  your f(x) must be continuous on the given interval.
Condition 2                    :  your f(x) must be differerentiable on that interval.

Are these conditions met?

3mar · Answer

The Mean Value Theorem states that if f(x) is $\color{red}{defined} $ and $\color{red}{continuous} $ on the interval [a,b] and $\color{red}{differentiable} $ on (a,b), then there is at least one number c in the interval (a,b) (that is a < c < b) such that:

$$\LARGE f '(\color{red}c)=\frac{ f(b)-f(a) }{ b-a }$$

mathmale · Answer

volpina:  Please involve yourself in this discussion.

volpina · Answer

So do i have to substitute into the Mean Value Theorm's equation and see if it meets the conditions?

mathmale · Answer

There's more to it than that.  FIRST, you determine whether or not the 2 conditions are satisfied.  Then, you calculate the quantities on each side of the given equation and set them equal to one another.  As before, your job is to determine whether nor not there is some value c between x=a and x=b at which this equation is true.

"What, if anything, does the Mean Value Theorem guarantee for the given function on the given interval?"

Reread the MVT and focus on the CONCLUSION.

mathmale · Answer

$$\LARGE f '(\color{red}c)=\frac{ f(b)-f(a) }{ b-a }$$

mathmale · Answer

You may as well go ahead and calculate all of the components in the above equation.  Given a and b values, evaluate f(a) and f(b).  Then  calculate $$\frac{ f(b)-f(a) }{ b-a }$$

mathmale · Answer

May I see your work and your result, please?

volpina · Answer

f(4)=(4)^2-2(4)+5=13
f(1)=(1)^2-2(1)+5=4
13-4/4-1
9/3
3

mathmale · Answer

OK.  Now calculate $$\frac{ f(b)-f(a) }{ b-a }$$      I see y ou've already given that a stab, but you haven't labeled your results.

mathmale · Answer

Is this what you meant?$$\frac{ f(b)-f(a) }{ b-a }=\frac{ 13-4 }{ 4-1 }?$$

mathmale · Answer

If so, calculate (and label) it.

volpina · Answer

$$\frac{ f(b)-f(a) }{ b-a }=\frac{ 13-4 }{ 4-1 }=3?$$

mathmale · Answer

Now please find the derivative of the given function.

f '(x)= ??

volpina · Answer

from the original function, it would 2x-2?

mathmale · Answer

Yes.  f '(x) = 2x-2.

Now replace both "x" with "c."  Write out your result.

mathmale · Answer

volpina?

volpina · Answer

is it 4?

mathmale · Answer

Your derivative was f '(x) = 2x - 2.  I'm asking you to throw out both "x" and replace them with 'c.'  please do that now.

f '(c) = ?

mathmale · Answer

We need to follow the road map of this Theorem closely.

mathmale · Answer

We have found [f(b)-f(a)]/[b-a] and now have to find f '(c).

mathmale · Answer

You do not yet have a value for c.  Instead, you have an equation for c.

mathmale · Answer

volpina, where have you been?

volpina · Answer

f '(c) = 2c - 2.