How Do I Solve This? g(x) = x3, -2 <= x <= 0 x2, 0 < x <= 2

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Abhishekdas

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Using Mean Value Theorem, find the value of c.

g(x) = x3, -2 <= x <= 0
x2, 0 < x <= 2

For which interval should I solve. I have done something like this. Kindly Check
IMG_0896.jpg
 
Abhishekdas said:
Using Mean Value Theorem, find the value of c.

g(x) = x3, -2 <= x <= 0
x2, 0 < x <= 2

For which interval should I solve. I have done something like this. Kindly Check
View attachment 9768
Click to expand...

I assume your statement of the MVT is something like this one: http://www.sosmath.com/calculus/diff/der11/der11.html

Note that there can be more than one value of "c", so "find the value of c" is inappropriate.

Also, the question should explicitly state the interval [a, b] in question.

But I would assume that, since they have defined a piecewise function over the interval [-2,2], that is probably the interval they are asking about -- not just either of the "pieces" of the domain, but the entire domain.

Try that.
 
mmm4444bot said:
:idea: Could the definite article be interpreted to denote that the given function is one-to-one?
Click to expand...

No, the function being one-to-one is irrelevant; the theorem still allows any number of answers.

I think the problem is just stated (or reported) sloppily. There are two answers, right?
 
Mvt

Subhotosh Khan said:
What is the "statement" of mean-value theorem?

What is "c" in that context?
Click to expand...
Mean Value Theorem states that if a function is continuous in the interval [a,b] and differentiable in (a,b), then there is a point c in the interval [a,b] such that

\(\displaystyle f'(c) = \dfrac{f(b) - f(a)}{b - a}\)
 
Last edited by a moderator:
OK

Dr.Peterson said:
I assume your statement of the MVT is something like this one: http://www.sosmath.com/calculus/diff/der11/der11.html

Note that there can be more than one value of "c", so "find the value of c" is inappropriate.

Also, the question should explicitly state the interval [a, b] in question.

But I would assume that, since they have defined a piecewise function over the interval [-2,2], that is probably the interval they are asking about -- not just either of the "pieces" of the domain, but the entire domain.

Try that.
Click to expand...

Ok. So, to find the absolute maximum and minimum, I should put the value of c in the according equation.
 
Abhishekdas said:
Mean Value Theorem states that if a function is continuous in the interval [a,b] and differentiable in (a,b), then there is a point c in the interval [a,b] such that

\(\displaystyle f'(c) = \dfrac{f(b) - f(a)}{b - a}\)
Click to expand...
Nobody has mentioned this yet, but you have not shown that your function is continuous on [a,b] and differentiable on (a,b). Showing this is the 1st step in using the MVT. Most students insist on not showing this but they're wrong.
 
Abhishekdas said:
… to find the absolute maximum and minimum, I should put the value of c in the according equation.
Click to expand...
Is this a different question? Are you asking about the absolute max/min for g(x) on the interval [-2,2]?

I wonder if before you were thinking c = critical number because we use critical numbers to find absolute max/min values.

We evaluate the function at the endpoints of the interval and also at any critical numbers in the interval.

Critical numbers are values of x where g'(x) = 0 or g'(x) is undefined.

Your function g(x) has one critical number in [-2, 2]. Call this x-value C. (I use big C here, to show that it's different from that little c in the MVT). Now evaluate the function at the endpoints and at C.

g(-2) = ?

g(C) = ?

g(2) = ?

The largest value is absolute max, and the smallest is the absolute min, on [-2,2].


Regarding the original question, I found two values for little c (as shown in the MVT link that Dr. Peterson posted). He suggested you try again, using the interval [-2,2]. Did you finish that part?

These values c1 and c2 give x-locations where tangent lines are parallel to the secant line (the line connecting the endpoints of g's graph). The lines all have same slope. That's a graphical description of MVT. Another description is that c1 and c2 are values where function g's instantaneous rate of change is the same as its average rate of change (over the given interval).

In this graph, function g is shown in red, the secant line is shown in blue (its slope is the average rate of change), and tangent lines are shown in green (their slope is the instantaneous rate of change at x=c1 and x=c2). The lines are parallel, showing their slopes are the same.

CLICK GRAPH TO ENLARGE

pwMV.jpg
 
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Dr.Peterson said:
I assume your statement of the MVT is something like this one: http://www.sosmath.com/calculus/diff/der11/der11.html

Note that there can be more than one value of "c", so "find the value of c" is inappropriate.

Also, the question should explicitly state the interval [a, b] in question.

But I would assume that, since they have defined a piecewise function over the interval [-2,2], that is probably the interval they are asking about -- not just either of the "pieces" of the domain, but the entire domain.

Try that.
Click to expand...
Thanks! I understood it now
 
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