The sum of all local minimum values of the fu | vedclass

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Question

(C) Given the function $f(x) = \begin{cases} 1-2x, & x  2 \end{cases}

Vedclass · Accepted Answer

$\frac{157}{72}$

Vedclass · Answer

(C) Given the function $f(x) = \begin{cases} 1-2x, & x  2 \end{cases}$.For $-1 \leq x \leq 2$,$f(x) = \frac{1}{3}(7+2|x|)$. This function has a local minimum at $x=0$ with value $f(0) = \frac{7}{3}$.For $x > 2$,$f(x) = \frac{11}{18}(x-4)(x-5) = \frac{11}{18}(x^2 - 9x + 20)$.To find the local minimum,we find the vertex of the parabola: $f'(x) = \frac{11}{18}(2x - 9) = 0 \implies x = \frac{9}{2} = 4.5$.The value at $x = 4.5$ is $f(4.5) = \frac{11}{18}(4.5-4)(4.5-5) = \frac{11}{18}(0.5)(-0.5) = \frac{11}{18} 	imes (-\frac{1}{4}) = -\frac{11}{72}$.Thus,the local minimum values are $\frac{7}{3}$ and $-\frac{11}{72}$.The sum of these values is $\frac{7}{3} - \frac{11}{72} = \frac{168 - 11}{72} = \frac{157}{72}$.

The sum of all local minimum values of the function $f(x) = \begin{cases} 1-2x, & x < -1 \\ \frac{1}{3}(7+2|x|), & -1 \leq x \leq 2 \\ \frac{11}{18}(x-4)(x-5), & x > 2 \end{cases}$ is:

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