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Calculus AB : Limits and Continuity

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Example Questions

Example Question #1 : Apply Intermediate Value Theorem

Can Intermediate Value Theorem be applied to the function f(x)=x^2 within the interval [-1,1] ?

Possible Answers:

No, because the function is not continuous

No, because f(-1)=f(1)

Yes, because the function crosses the axis within the interval [-1,1]

Yes, because the function f(x)=x^2 has a root at (0,0)

Correct answer:

No, because f(-1)=f(1)

Explanation:

The required conditions for Intermediate Value Theorem include the function must be continuous and f(a) cannot equal f(b) . While there is a root at (0,0) for this particular continuous function, this cannot be shown using Intermediate Value Theorem. The function does not cross the axis, thus eliminating that particular answer choice. The correct answer is “No, because f(-1)=f(1) .” Since one of the conditions for Intermediate Value Theorem is that f(a) cannot equal f(b) , by graphing f(x)=x^2 we can see that this requirement is not met.

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Example Question #121 : Limits And Continuity

What can be concluded by using Intermediate Value Theorem for the function f(x)=(x-2)^3 on the interval [1,3] ?

Possible Answers:

There is a root for this polynomial at x=2

There are two roots on this polynomial located between x=1 and x=3

The requirements for Intermediate Value Theorem are not met

There is a root for this polynomial located between x=1 and x=3

Correct answer:

There is a root for this polynomial located between x=1 and x=3

Explanation:

his function is continuous (as it is a polynomial) and $f(1)\neq f(3)$ ; therefore, the required conditions for Intermediate Value Theorem are met. While there is a root at x=2 for this function (as can be seen by graphing the polynomial), Intermediate Value Theorem does not state where this root will be exactly, nor does it state how many roots there might be. Thus, the conclusion that can be made by IVT is that there is a root for this polynomial located somewhere between x=1 and x=3 .

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Example Question #122 : Limits And Continuity

Let f(x)=x^2 . Is there a number between and such that f(c)=7 ?

Possible Answers:

No, no number c such that f(c)=7 exists

Yes, as shown by Intermediate Value Theorem

No, Intermediate Value Theorem cannot determine the exact value of f(c)

Yes, as shown by the Fundamental Theorem of Calculus

Correct answer:

Yes, as shown by Intermediate Value Theorem

Explanation:

First, determine the values of the function at the bounds. This will allow the correct implementation of the Intermediate Value Theorem.

f(x)=x^2

f(2)=2^2=4

f(3)=3^2=9

Because the problem asks to analyze the interval [2,3] and f(2)=4<f(3)=9 , there must be a value f(a)<f(c)<f(b) , with a<c<b . Because f(2)=4<f(c)=7<f(3)=9 , by Intermediate Value Theorem there should be a number between and that satisfies the required conditions. Therefore, “Yes, as shown by Intermediate Value Theorem” is the correct answer.

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Example Question #123 : Limits And Continuity

Assume f(x) is continuous on the interval [-1,1] and has the values listed in the table below. Which of the following values of guarantees that f(x) has at least two roots?

Q10 table

Possible Answers:

Correct answer:

Explanation:

If b=0 , only one root can be guaranteed (at x=0 ).

If b= -1 , then Intermediate Value Theorem can be applied twice, for [-1,0] and [0,1] .

This is true because for continuous functions, Intermediate Value Theorem states that a change in sign (ex: from positive to negative) of the function within an interval suggests a root (where the function crosses the axis) at some point within that interval.

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Calculus AB : Limits and Continuity

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

Example Question #1 : Apply Intermediate Value Theorem

Example Question #121 : Limits And Continuity

Example Question #122 : Limits And Continuity

Example Question #123 : Limits And Continuity

All Calculus AB Resources

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