Question

The sum of the absolute minimum and the absolute maximum values of the function ƒ(x) = |3x – x² + 2| – x in the interval [–1, 2] is

• A. \(\frac{\sqrt17+3}{2}\)
• B. \(\frac{\sqrt{17}+5}{2}\)
• C. 5
• D. \(\frac{9-\sqrt{17}}{2}\)

Answer 1

The Correct Option is A

The correct answer is (A) : \(\frac{\sqrt17+3}{2}\)
ƒ(x) = |x²– 3x – 2| – x
\(∀x∈[−1,2]\)
\(f(x) = \begin{cases}        x^2 - 4x - 2 & \text{if } -1 \leq x < \frac{3 - \sqrt{17}}{2} \\       -x^2 + 2x + 2 & \text{if } \frac{3 - \sqrt{17}}{2} \leq x \leq 2 \end{cases}\)

Fig.

\(ƒ(x)_{max} = 3\)
\(ƒ(x)_{min}=ƒ(\frac{3−\sqrt{17}}{2})\)
\(=\frac{\sqrt{17}-3}{2}\)