MHT CET2008MathematicsApplication of Derivatives

The maximum value of function x³-12 x²+36 x+17 in the interval [1,10] is

Options

A17
B177
C77
DNone of these

Correct answer

B. 177

Step-by-step solution

Let f(x)=x³-12 x²+36 x+17 f^ (x)=3 x²-24 x+36=0 For maxima, put f^ (x)=0 3 x²-24 x+36=0 (x-2)(x-6)=0 x=2,6 Again, f^ (x)=6 x-24 is negative at x=2 So that, f(6)=17, f(2)=49 At the end points, f(1)=42, f(10)=177 So that, f(x) has its maximum value 177 .

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