The sum of absolute maximum and absolute mini | vedclass

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(B) Given $f(x) = |2x^2 + 3x - 2| + \sin x \cos x = |(2x-1)(x+2)| + \frac{1}{2} \sin(2x)$.In the interval $[0, 1]$,$2x-1$ changes sign at $x = 1/2$. S

Vedclass · Accepted Answer

$3+\frac{1}{2}(1+2 \cos (1)) \sin (1)$

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(B) Given $f(x) = |2x^2 + 3x - 2| + \sin x \cos x = |(2x-1)(x+2)| + \frac{1}{2} \sin(2x)$.In the interval $[0, 1]$,$2x-1$ changes sign at $x = 1/2$. Since $x+2 > 0$ for $x \in [0, 1]$,we have:$f(x) = \begin{cases} -(2x^2 + 3x - 2) + \frac{1}{2} \sin(2x), & 0 \leq x For $0 \leq x  3$ and $\cos(2x) \leq 1$,$f'(x) For $1/2  0$,so $f(x)$ is strictly increasing.Thus,the absolute minimum occurs at $x = 1/2$: $f(1/2) = |0| + \frac{1}{2} \sin(1) = \frac{1}{2} \sin(1)$.The absolute maximum occurs at the endpoints $x=0$ or $x=1$.$f(0) = |-2| + 0 = 2$.$f(1) = |2+3-2| + \frac{1}{2} \sin(2) = 3 + \frac{1}{2} \sin(2)$.Comparing $f(0)=2$ and $f(1)=3 + \frac{1}{2} \sin(2)$,since $\sin(2) > 0$,$f(1) > f(0)$.Absolute maximum is $3 + \frac{1}{2} \sin(2) = 3 + \sin(1) \cos(1)$.Sum = $f(1/2) + f(1) = \frac{1}{2} \sin(1) + 3 + \sin(1) \cos(1) = 3 + \frac{1}{2} \sin(1) (1 + 2 \cos(1))$.

The sum of absolute maximum and absolute minimum values of the function $f(x)=|2 x^{2}+3 x-2|+\sin x \cos x$ in the interval $[0,1]$ is

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