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# Justification with the mean value theorem: equation

Example justifying use of mean value theorem (where function is defined with an equation).

## Want to join the conversation?

• Can we get more explanation on
"All Rational functions are continuous and differentiable in it's domain"
• A rational function is one polynomial divided by another. Polynomials are differentiable everywhere, so we can always differentiate a rational function using the quotient rule, unless the denominator is zero at that point.

But if the denominator is 0 at some point a, then a is not in the domain of the rational function. So for every point that is in the domain, we can differentiate the function normally.
• At , if all rational functions are continuous and differentiable at every point in its domain, why is g(x) not continuous at x = 0? Isn't x = 0 a part of its domain?
• The domain of a function is the set of input values for which the function is defined.
𝑔(𝑥) = 1∕𝑥 is not defined for 𝑥 = 0 (due to division by zero), and thereby 𝑥 = 0 is not part of its domain.
• for the given information I solved it like this,
1.
g(x)=1/x
g'(x)=-1/x^2
given that,
g'(x)=1/2
-1/x^2=1/2
x=(-2)^(1/2) which is not the element of (-1,2)
therefore, No

2.
g(x)=1/x
g'(x)=-1/x^2
given that,
g'(c)=-1/2
-1/x^2=1/2
x=(2)^(1/2) which is the element of (1,2)
therefore, Yes

Is this okey?
• I think you should have checked if the function was continuous & differentiable first, but otherwise I would say this works
(1 vote)
• at how do we find out that g(2) is 1/2?
• The function is given to be g(x) = 1/x, so g(2) = 1/2
• Is the solution sqrt of 2??
• Where does the equation at come from and how did Sal know it was equal to fprime of c?
• Because that's what the Mean Value Theorem states: If a function f(x) is differentiable on an interval [a,b], then there is a point c in [a, b] where f'(c)=[f(b)-f(a)]/(b-a).

This is just an application of the mean value theorem with a=1, b=2.
• Why is differentiability necessary for the validity of the Mean Value Theorem?
• If the function is not differentiable, (i.e. it is discontinuous and/or contains a sharp curve), we cannot guarantee that there exists a point in the interval whose derivative is equal to the slope of the line passing through the points (a,f(a)) and (b,f(b)).
(1 vote)
• what if the equation is y=e^x
• It is continuous as well as differentiable for all x. So, MVT can be applied.
(1 vote)
• Quick question (about part 2 in the video):

So if the f'(c)=-1/2, the only value of x (I can find) that makes that true is x= 4 which is outside our interval (1,2)...