If f(x)=|x-5|+|x+5|+|x-4|+|x+4|,then frac{f^{ | vedclass

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(A) Given the function $f(x)=|x-5|+|x+5|+|x-4|+|x+4|$.We can define $f(x)$ in different intervals:$f(x) = \begin{cases} -4x & x \leq -5 \ -2x+10 & -5

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(A) Given the function $f(x)=|x-5|+|x+5|+|x-4|+|x+4|$.We can define $f(x)$ in different intervals:$f(x) = \begin{cases} -4x & x \leq -5 \ -2x+10 & -5  5 \end{cases}$Now,the derivative $f^{\prime}(x)$ is:$f^{\prime}(x) = \begin{cases} -4 & x  5 \end{cases}$Evaluating the required values:$f^{\prime}(1) = 0$ (since $-4 $f^{\prime}(-6) = -4$ (since $-6 $f^{\prime}(-1) = 0$ (since $-4 $f^{\prime}(6) = 4$ (since $6 > 5$)Substituting these into the expression:$\frac{f^{\prime}(1)-f^{\prime}(-6)}{f^{\prime}(-1)+f^{\prime}(6)} = \frac{0-(-4)}{0+4} = \frac{4}{4} = 1$.

If $f(x)=|x-5|+|x+5|+|x-4|+|x+4|$,then $\frac{f^{\prime}(1)-f^{\prime}(-6)}{f^{\prime}(-1)+f^{\prime}(6)}=$

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