The sum of the maximum and minimum values of | vedclass

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(B) Given $f(x) = |5x - 7| + [x^2 + 2x]$.For $x \in [\frac{5}{4}, 2]$,we analyze the function.At $x = \frac{5}{4} = 1.25$,$f(1.25) = |5(1.25) - 7| + [

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(B) Given $f(x) = |5x - 7| + [x^2 + 2x]$.For $x \in [\frac{5}{4}, 2]$,we analyze the function.At $x = \frac{5}{4} = 1.25$,$f(1.25) = |5(1.25) - 7| + [1.25^2 + 2(1.25)] = |6.25 - 7| + [1.5625 + 2.5] = |-0.75| + [4.0625] = 0.75 + 4 = 4.75$.At $x = \frac{7}{5} = 1.4$,$f(1.4) = |5(1.4) - 7| + [1.4^2 + 2(1.4)] = 0 + [1.96 + 2.8] = [4.76] = 4$.Since both $|5x - 7|$ and $x^2 + 2x$ are increasing for $x > 1.4$,the function $f(x)$ is increasing on $[1.4, 2]$.At $x = 2$,$f(2) = |5(2) - 7| + [2^2 + 2(2)] = |10 - 7| + [4 + 4] = 3 + 8 = 11$.The minimum value is $4$ and the maximum value is $11$.The sum of the maximum and minimum values is $4 + 11 = 15$.

The sum of the maximum and minimum values of the function $f(x) = |5x - 7| + [x^2 + 2x]$ in the interval $[\frac{5}{4}, 2]$,where $[t]$ denotes the greatest integer function $\leq t$,is:

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