In the interval (7, infty),the function f(x) | vedclass

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(A) Given the function $f(x) = |x-5| + 2|x-7|$. For the interval $(7, \infty)$,we have $x > 7$. Since $x > 7$,it follows that $x > 5$ and $x > 7$. The

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increasing function

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(A) Given the function $f(x) = |x-5| + 2|x-7|$. For the interval $(7, \infty)$,we have $x > 7$. Since $x > 7$,it follows that $x > 5$ and $x > 7$. Therefore,$|x-5| = x-5$ and $|x-7| = x-7$. Substituting these into the function: $f(x) = (x-5) + 2(x-7)$ $f(x) = x - 5 + 2x - 14$ $f(x) = 3x - 19$. Now,find the derivative $f'(x)$: $f'(x) = \frac{d}{dx}(3x - 19) = 3$. Since $f'(x) = 3 > 0$ for all $x$ in the interval $(7, \infty)$,the function is an increasing function.

In the interval $(7, \infty)$,the function $f(x) = |x-5| + 2|x-7|$ is:

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