If C1,C2 are the values of x,(C1<C2) for which LMVT holds for the function f(x)=x3 on the interval [−3,3], then the value of 4C21+7C22 is equal to
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Solution
f(x)=x3,x∈[−3,3] f is continuous in [−3,3] and differentiable on (−3,3)
So, from LMVT, f′(c)=f(3)−f(−3)3−(−3) ⇒3c2=9 ⇒c=±√3∈(−3,3) ∴C1=−√3,C2=√3 ⇒4C21+7C22=33