TS EAMCET · Maths · Application of Derivatives
The sum of the maximum and minimum values of the function \(f(x)=\frac{x^2-x+1}{x^2+x+1}\) is
- A \(\frac{17}{4}\)
- B \(\frac{5}{2}\)
- C \(\frac{10}{3}\)
- D 0
Answer & Solution
Correct Answer
(C) \(\frac{10}{3}\)
Step-by-step Solution
Detailed explanation
\(\begin{gathered}f(x)=\frac{x^2-x+1}{x^2+x+1} \\ f^{\prime}(x)=\frac{\left(x^2+x+1\right)(2 x-1)-\left(x^2-x+1\right)(2 x+1)}{\left(x^2+x+1\right)^2}\end{gathered}\) Critical points are \(x=1, x=-1\) \(f(x)\) is maximum at \(x=-1\) \(f(-1)=\frac{1+1+1}{1-1+1}=3\) \(f(x)\) is…
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