The function f(x) = |px - q| + r|x|, x in ( - | vedclass

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(C) Given the function $f(x) = |px - q| + r|x|$.We analyze the critical points where the expression inside the absolute value becomes zero: $x = q/p$

Vedclass · Accepted Answer

$r 
eq p$

Vedclass · Answer

(C) Given the function $f(x) = |px - q| + r|x|$.We analyze the critical points where the expression inside the absolute value becomes zero: $x = q/p$ and $x = 0$.Case $1$: If $x  0$,the slope $-(p+r)$ is negative,so $f(x)$ is decreasing.Case $2$: If $0 \le x \le q/p$,$f(x) = -(px - q) + rx = (r-p)x + q$.Case $3$: If $x > q/p$,$f(x) = (px - q) + rx = (p+r)x - q$. Since $p, r > 0$,the slope $(p+r)$ is positive,so $f(x)$ is increasing.For the function to have a unique minimum,the slope in the interval $[0, q/p]$ must be zero,which implies $r - p = 0$,or $r = p$.However,if $r = p$,the function becomes $f(x) = |px - q| + p|x|$. At $x = q/p$,$f(q/p) = |p(q/p) - q| + p|q/p| = 0 + q = q$. At $x = 0$,$f(0) = |0 - q| + p(0) = q$.If $p = q = r$,then $f(x) = |px - p| + p|x| = p(|x - 1| + |x|)$. The minimum value is $p$ for all $x \in [0, 1]$.Re-evaluating the condition for a single point: The function $f(x) = |px - q| + r|x|$ has a unique minimum at $x = q/p$ if the slope changes from negative to positive. This occurs when $r 
eq p$ and $r 
eq -p$ (which is always true as $r, p > 0$). Specifically,if $r > p$,the minimum is at $x = 0$. If $r < p$,the minimum is at $x = q/p$. If $r = p$,the function is constant on $[0, q/p]$. Thus,the condition for a unique minimum is $r 
eq p$.

The function $f(x) = |px - q| + r|x|,$ $x \in ( - \infty, \infty )$ where $p > 0, q > 0, r > 0$ assumes its minimum value only at one point if

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