Question

Show that the equation x^3-15x+3=0 has exactly one real root in [-2,2].

I'm not sure of the steps needed to solve this? I know it involves Rolles Theorem?

Answer 1

it's actually the intermediate value theorem

Answer 2

if you can show there is a sign change in the y values for f(-2) to f(2), then there has to be a root somewhere between x = -2 and x = 2

this is assuming f(x) is continuous

Answer 3

actually nevermind, that's something slightly similar

take a look at this page

http://tutorial.math.lamar.edu/Classes/CalcI/MeanValueTheorem.aspx

Answer 4

look at Example 1

Answer 5

http://www.wolframalpha.com/input/?i=plot+x^3-15x%2B3+from+x%3D-2+to+x%3D2

Answer 6

Before you attempt to apply the Intermediate Value Theorem, you need to check to ensure that the conditions of this Theorem are satisfied. have you done that? For example, the function must be continuous on the open interval (-2,2).

Note that "has a root" here signifies the same thing as "crosses the x-axis" at some value of x between -2 and 2.

Answer 7

I plugged in -2 and 2 and I got a negative and positive number, 25 and -19, does this satisfy the conditions of the IVT? And means its continuous? @mathmale

Answer 8

@jim_thompson5910 Thank you! that example helps I am just having a hard time understanding Rolles Theorem and how it applies

Answer 9

I'll follow the template given in the example

f(x) = x^3-15x+3 is given

let's find f(-2) and f(2)

f(x) = x^3-15x+3

f(-2) = (-2)^3 - 15(-2) + 3

f(-2) = 25

------------

f(x) = x^3-15x+3

f(2) = (2)^3 - 15(2) + 3

f(2) = -19

------------

the sign change from + to - when you go from f(-2) to f(2) implies that f(x) = 0 at least once between x = -2 and x = 2


Now let's take the derivative to get f ' (x) = 3x^2 - 15


solve f ' (x) = 0

f ' (x) = 0

3x^2 - 15 = 0

3(x^2 - 5) = 0

x^2 - 5 = 0

x^2 = 5

x = sqrt(5) or x = -sqrt(5)

x = 2.236 or x = -2.236


Notice how x = 0 is between these two roots and how f(0) = -15. So the derivative is negative between x = -2.236 and x = 2.236


In other words, f ' (x) < 0 when -sqrt(5) < x < sqrt(5) or -2.236 < x < 2.236


This negative derivative throughout this interval means that the function is decreasing throughout the interval.


So effectively what happens is that the function f(x) passes through the root, wherever it is, exactly once. It passes downward because it's decreasing here. It does NOT come back up to intersect a second time because it's always decreasing on this specific interval.

Answer 10

f(-2) = 25 ---> (-2,25) is on the graph

Answer 11

f(2) = -19 ---> (2,-19) is on the graph