MHT CET · Maths · Application of Derivatives
The maximum value of function \(x^{3}-12 x^{2}+36 x+17\) in the interval \([1,10]\) is
- A 17
- B 177
- C 77
- D None of these
Answer & Solution
Correct Answer
(B) 177
Step-by-step Solution
Detailed explanation
Let \(f(x)=x^{3}-12 x^{2}+36 x+17\)
\(\therefore f^{\prime}(x)=3 x^{2}-24 x+36=0\)
For maxima, put \(f^{\prime}(x)=0\) \(\Rightarrow \quad 3 x^{2}-24 x+36=0\)
\(\Rightarrow \quad(x-2)(x-6)=0\)
\(\Rightarrow \quad x=2,6\)
Again, \(f^{\prime \prime}(x)=6 x-24\) 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 .
\(\therefore f^{\prime}(x)=3 x^{2}-24 x+36=0\)
For maxima, put \(f^{\prime}(x)=0\) \(\Rightarrow \quad 3 x^{2}-24 x+36=0\)
\(\Rightarrow \quad(x-2)(x-6)=0\)
\(\Rightarrow \quad x=2,6\)
Again, \(f^{\prime \prime}(x)=6 x-24\) 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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