How do you determine all values of c that satisfy the mean value theorem on the interval [-6, 7] for #f(x)=2x^3–9x^2–108x+2#?

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Answer 1 · 2016-08-19T17:51:11

#c_1=5.38, &, c_2=-2.38 in (-6,7)#

Here, #f : [-6,7] rarr RR,# defined by, #f(x)=2x^3-9x^2-108x+2#.

#f# is a cubic poly., &, we know that all polys. are cot. & derivable on #RR#.

#:. f# is cont. on #[-6,7]#......(MVT1), &, derivable #(-6,7)#......(MVT2)

Thus, #f# satisfies the conds. (MVT1) and (MVT2) of the Mean Value Theorem.

Therefore, #EE# some #c in (-6,7), s.t., f'(c)=(f(7)-f(-6))/(7-(-6))...(MVT)#.

#f(x)=2x^3-9x^2-108x+2 rArr f'(c)=6c^2-18c-108#

Also, #(f(b)-f(a))/(b-a)=(2(b^3-a^3)-9(b^2-a^2)-108(b-a))/(b-a)#

#=2(b^2+ba+a^2)-9(b+a)-108#

#:. (f(7)-f(-6))/(7-(-6))=2(49-42+36)-9(7-6)-108=86-9-108=-31#

Therefore, by #(MVT), 6c^2-18c-108=-31#, or,

#6c^2-18c-77=0#

#:. c=(18+-sqrt(324+1848))/12=(18+-sqrt2172)/12=(18+-46.60)/12#

#:. c_1=5.38, &, c_2=-2.38 in (-6,7)#