Given the function #f(x)= abs((x^2-12)(x^2+4))#, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [-2,3] and find the c?

Answer 1

Here's a sketch or outline of the solution.

#f# is the composition of the absolute value and a polynomial, so #f# is continuous at every real number. In particular, #f# is continuous on #[-2,3]#. So #f# satisfies the first hypothesis on this interval..
In general #absu# is differentiable expect where #u# changes signs.
#(x^2 -12)# changes signs at #x= +- sqrt2#.
#(x^2+4)# deos not changes signs on the real numbers.
So #f# is non-differentiable ony at #+- sqrt12# .
Since #3 < sqrt12 < 4#, neither of #+- sqrt12# is in #(-2,3)#
So #f# is differentiable on #(-2,3)# And #f# satisfies the second hypothesis on this interval..
The Mean Value Theorem now assures us that there is a #c# in #(-2,3)# with #f'(c) = (f(3)-f(-2))/(3-(-2))#
Actually finding the #c# is tedious.
On the interval #[-2,3]#, the values of #(x^2-12)(x^2+4)# are negative, so
#f(x) = -(x^2-12)(x^2+4) = -x^4+8x^2+48#
And #f'(x) = -4x^3+16x#
#f(3) = 39# and #f(-2) = 64#, so we need to solve
#-4x^3 +16x = (39-64)/(3-(-2)#
#-4x^3+16x = -7#
#4x^3-16x-7=0#
Now use whatever techniques you have available to solve this third degree equation. (There are two solutions in #(-2,3)# and another outside the interval.)
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To determine whether the function satisfies the hypotheses of the Mean Value Theorem (MVT) on the interval ([-2,3]), you need to check if the function is continuous on the closed interval and differentiable on the open interval.

  1. Continuity: The function (f(x) = \text{abs}((x^2 - 12)(x^2 + 4))) is continuous everywhere because it is a composition of continuous functions.

  2. Differentiability: To check differentiability, you need to ensure that the function is differentiable on the open interval (-2, 3). Differentiability of a function is determined by the differentiability of its components.

    (f(x) = \text{abs}((x^2 - 12)(x^2 + 4))) is differentiable at every point where its components (g(x) = (x^2 - 12)) and (h(x) = (x^2 + 4)) are differentiable. The functions (g(x)) and (h(x)) are polynomials and hence are differentiable everywhere.

Since the function (f(x)) is continuous on the closed interval [-2, 3] and differentiable on the open interval (-2, 3), it satisfies the hypotheses of the Mean Value Theorem on the interval [-2,3].

To find (c), we use the Mean Value Theorem formula:

[f'(c) = \frac{f(3) - f(-2)}{3 - (-2)}]

Calculate (f(3)) and (f(-2)), then find the derivative of (f(x)) and solve for (c).

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7