Question #ba71a

1 Answer
Sridhar V.
Feb 8, 2018

Rolle's Theorem is satisfied as there are two points at

#color(blue)(x=1.868517," " x = -0.53518#

where #color(green)(f'(x) = 0#

Explanation:

Rolle's Theorem states that under certain conditions an extreme value is guaranteed to lie in the interior of the closed interval.

Rolle's Theorem

Conditions:

  1. The function #color(blue)(f(x)# must be continuous on the closed interval [a, b].

  2. The function f(x) must be differentiable on the open interval (a, b).

  3. f(a) = f(b)

If these three conditions are satisfied then there is at least one number in the open interval (a, b) such that the derivative of f(x) is zero.

Given:

#color(brown)(f(x) = x^3-2x^2-3x#

Observe that #color(brown)(f(x) = x^3-2x^2-3x# is a Cubic Function.

We will now find the x-intercepts of #f(x)#

#color(green)(f(x) = x^3-2x^2-3x#

and show that

#f'(x) = 0 # at some points between the x-intercepts

We have

#f(x) = x^3-2x^2-3x#

#f(x) = x(x^2-2x-3)#

We can write this expression as

#f(x) = x(x^2+x-3x-3)#

#f(x) = x[x(x+1)-3(x+1)]#

#rArr x*(x+1)*(x-3)#

Hence the solutions to our Cubic Function are

#color(blue)(x=0; x= (-1); x= 3#

Please refer to the image of the graph below that visually presents our findings.

enter image source here

Next

we will differentiate #color(green)(f(x) = x^3-2x^2-3x#

We will find

#d/dx [x^3 - 2x^2 - 3x]#

#rArr 3x^2-4x-3#

#:. f'(x) = 3x^2-4x-3#

Next

we will set #color(brown)(f'(x) = 0#

i.e., #3x^2-4x-3 = 0#

This is a quadratic equation

General Form of a quadratic equation is

#color(red)(ax^2+bx+c=0#

Consider our quadratic equation

#3x^2-4x-3 = 0#

Note that

#color(blue)(a=3; b=(-4) and c=(-3)#

We will use the following quadratic formula to simplify:

#x=[ -b +- sqrt(b^2 - 4ac)]/(2a)#

#x=[ -(-4) +- sqrt((-4)^2 - 4*3*(-3))]/(2*3)#

#x=[4+-sqrt(52)]/6#

Hence our roots are

#x=(2/3)+(1/3)*sqrt(13) " "(or)" "#

#x = (2/3)+(-1/3)*sqrt(13)#

#color(blue)(rArr x = 1.868517 or x = - 0.53518#

Rolle's Theorem is satisfied as there are two points at

#color(blue)(x=1.868517," " x = -0.53518#

where #color(green)(f'(x) = 0#

Please refer to the image of the graph below that visually presents our findings.

enter image source here

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