How do you verify that the function #f(x)=x^(3)-x^(2)-12x+4# satisfies the three hypotheses of Rolle's Theorem on the given interval [0,4] and then find all numbers c that satisfy the conclusion of Rolle's Theorem?

Question

63276 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-06T08:25:46

c=#(2+sqrt148)/6# = 2.36...

Here f(0)= 4 and f(4)=4 Hence according to the Rolle's theorem there should be at least one c for which f'(c)=0.

To find c, get #f'(x)=3x^2-2x-12# and equate it to 0

#3x^2-2x-12#=0 and solve for x using quadratic formula.

x=# (2+-sqrt (4+144))/6#. This gives the desired c=#(2+sqrt 148)/6# in the interval[0,4]

Answer 2 · 2015-09-06T13:15:31

Check the hypotheses, then solve #f'(x) = 0# to find:

#c = (1+sqrt(37))/3#

Let's check the hypotheses of Rolle's Theorem:

(1) #f(x)# is continuous on #[0, 4]#

#f(x)# is a polynomial in #x#, so continuous on any interval.

(2) #f(x)# is differentiable on #(0, 4)#

#f'(x) = 3x^2-2x-12#

(3) #f(0) = f(4)#

#f(0) = 0 - 0 - 0 + 4 = 4#
#f(4) = 64 - 16 - 48 + 4 = 4#

So according to Rolle's Theorem #f'(x) = 0# for some #x in (0, 4)#

#f'(x) = 3x^2-2x-12#

The roots of #f'(x) = 0# are given by the quadratic formula as:

#x = (2+-sqrt(2^2-(4xx3xx-12))) / (2*3)#

#=(2+-sqrt(148))/6#

#=(1+-sqrt(37))/3#

Now #sqrt(37) > 1#, so #(1-sqrt(37))/3 < 0# is outside our interval.

This leaves the solution #c = (1+sqrt(37))/3#