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INSTRUCTORS Raffi Hovasapian John Zhu
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Lecture Comments (13)

1 answer

Last reply by: Marcelo Quijano
Fri Nov 7, 2014 8:08 PM

Post by Marcelo Quijano on November 7, 2014

Can someone explain to me how he messed up on example 2?

2 answers

Last reply by: Chonglin Xu
Thu Jan 1, 2015 1:12 AM

Post by BAILIN ZHANG on February 5, 2014

Example 1 is not correct.

0 answers

Post by Jimmy Wu on October 20, 2013

Remember, both points are a straight line. It can either be a Slope of 0/1 or it can also be of the same value.

0 answers

Post by Jimmy Wu on October 20, 2013

Example 3 can be applied. A must equal to B which does.

0 answers

Post by Sahil Gupta on January 20, 2013

You messed up in example 1 and 2.

4 answers

Last reply by: Akshay Tiwary
Sat Apr 27, 2013 10:17 AM

Post by abbas esmailzadeh on November 3, 2012

example 1:f(x)=x^2-x how f'(x)=4x-4

Rolles Theorem

  • There exists a point c, between a and b where the slope is 0.
  • Applying Rolle’s Theorem
    • Check conditions
    • Take derivative of function
    • Set derivative function = 0
    • Solve for c value
  • Most often straight forward. Just check the conditions and apply the method

Rolles Theorem

Use Rolle's Theorem to show that f′(c) = 0 for some c in the interval [0, 4] with f(x) = 4x2 − 16x + 4
  • Polynomial, so it's continuous
  • f′(x) = 8x − 16 defined for all x
  • 4x2 − 16x + 4 = 0
  • x = [(16 ±√{(−16)2 − 4(4)(4)})/2(4)]
  • x = [(16 ±√{256 − 64})/8]
  • x = 2 ±[(√{192})/8]
  • x = 2 ±[(√{64 * 3})/8]
  • x = 2 ±√3 Both solutions are within the interval
Yes. There exists some c in the interval [0,4] such that f′(c) = 0.
Find the value of x such that f′(x) = 0 for f(x) = 4x2 − 16x + 4
  • f′(x) = 8x − 16
  • 8x − 16 = 0
  • 8x = 16
  • x = [16/8]
f′(2) = 0
Use Rolle's Theorem to show that f′(c) = 0 for some c in the interval [−1, 1] with f(x) = |x| − 1
  • We can rewrite this as f(x) = √{x2} − 1
  • The function is continuous on the interval
  • f′(x) = [1/2] [1/(√{x2})] [d/dx] x2
  • f′(x) = [x/(√{x2})] Removable discontinuity at x = 0
  • limx → 0+ f′(x) ≠ limx → 0 f′(x)
The function is not differentiable at x = 0, which is in the interval. Rolle's Theorem cannot be applied.
Confirm the Rolle's Theorem conditions on the interval [−1, 1] with f(x) = x − 1
  • Polynomial, so it's continuous
  • f′(x) = 1
  • Differentiable on that interval
  • Let's check the endpoints
  • f(−1) = −2
  • f(1) = 0
  • The endpoints don't match. They don't necessarily have to equal zero, but there must be two equal values of y within that interval. Rolle's Theorem is only a special case of the Mean Value Theorem, which is covered in the next lesson
The conditions for Rolle's Theorem are not met.
Use Rolle's Theorem to show that f′(c) = 0 for some c in the interval [0, π] with f(x) = sinx
  • f(x) = sinx
  • sinx is defined for all x. Continuous
  • f′(x) = cosx
  • cosx is defined for all x and has no edges or kinks in its graph. Differentiable.
  • sinx = 0
  • x = 0, π
Yes. There exists some c in the interval [0, π] such that f′(c) = 0.
Find the value of x in the interval [0, π] such that f′(x) = 0 for f(x) = sinx
  • f′(x) = cosx
  • f′([(π)/2]) = 0
f′([(π)/2]) = 0
Use Rolle's Theorem to show that f′(c) = 0 for some c in the interval [0, 4] with f(x) = x3 − 12x
  • Polynomial function, so it's continuous everywhere
  • f′(x) = 3x2 − 12 continuous everywhere
  • x3 − 12x = 0
  • x(x2 − 12) = 0
  • x(x + √{12})(x − √{12}) = 0
  • Three solutions, x = 0, −√{12}, √{12}, and two of these are within the interval
Yes. There exists some c in the interval [0, 4] such that f′(c) = 0.
Use Rolle's Theorem to show that f′(c) = 0 for some c in the interval [−2, 2] with f(x) = ex
  • ex is a continuous function so it's continuous function. In fact, it can be expressed as a polynomial function
  • f′(x) = ex defined everywhere. In fact, it can be expressed as a polynomial function
  • ex ≠ 0
  • Let's test the end-points
  • f(−2) = e−2
  • f(2) = e2
  • f(2) ≠ f(−2)
Rolle's Theorem conditions not met.
Find x such that f′(x) = 0 where f(x) = x2
  • f′(x) = 2x
  • f′(0) = 0
When x = 0, f′(x) = 0. Rolle's Theorem can be used to prove that a solution in an interval exists, but it doesn't necessarily prove there is no solution. In truth, the same
Use Rolle's Theorem to show that f′(c) = 0 for some c in the interval [0, 1] with f(x) = x4 − x2
  • Polynomial function, so it's continuous
  • f′(x) = 4x3 − 2x is defined everywhere
  • x4 − x2 = 0
  • x2(x2 − 1) = 0
  • x = −1, 0, 1
Yes. There exists some c in the interval [0, 1] such that f′(c) = 0.

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Rolles Theorem

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Rolle's Theorem 0:07
    • Conditions
    • Summary
  • Example 1 1:09
  • Example 2 3:08
  • Example 3 4:48