Finding Absolute Maximum and Minimum Values

Find the absolute maximum value and absolute minimum value of the function \[ f(x)=\left\{ {{ \begin{array}{@{}l@{\qquad}l@{\quad}l} {2x-1} & \text{if} & {0\leq x\leq 2} \\ {x^{2}-5x+9} & \text{if} & {2<x\leq 3} \end{array} }}\right. \]

Solution The function \(f\) is continuous on the closed interval \([0,3]\). (You should verify this.) To find the absolute maximum value and absolute minimum value, we follow the three-step procedure.

Step 1 Find the critical numbers in the open interval \((0,3)\):

On the open interval \((0, 2)\): \(f(x)= 2x-1\) and \(f^\prime (x) =2\). Since \(f^\prime (x) \neq 0\) on the interval \((0, 2) ,\) there are no critical numbers in \((0, 2)\).

On the open interval \((2, 3)\): \(f(x)=x^{2}-5x+9\) and \(f^\prime (x) =2x - 5\). Solving \(f^\prime (x) =2x-5=0\), we find \(x=\dfrac{5}{2}\). Since \(\dfrac{5}{2}\) is in the interval \((2, 3)\), \(\dfrac{5}{2}\) is a critical number.

At \(x=2\), the rule for \(f\) changes, so we investigate the one-sided limits of \(\dfrac{f(x) -f(2) }{x-2}.\) \[ \begin{eqnarray*} \lim\limits_{x\rightarrow 2^{-}}\dfrac{f(x) -f(2) }{ x-2} &=& \lim\limits_{x\rightarrow 2^{-}}\dfrac{(2x-1) -3}{x-2} =\lim\limits_{x\rightarrow 2^{-}}\dfrac{2x-4}{x-2}\\[3pt] &=&\lim\limits_{x\rightarrow 2^{-}}\dfrac{2(x-2) }{x-2}=2 \\[3pt] \lim\limits_{x\rightarrow 2^{+}}\dfrac{f(x) -f(2) }{ x-2} &=& \lim\limits_{x\rightarrow 2^{+}}\dfrac{(x^{2}-5x+9) -3}{x-2} =\lim\limits_{x\rightarrow 2^{+}}\dfrac{x^{2}-5x+6}{x-2}\\[3pt] &=&\lim\limits_{x \rightarrow 2^{+}}\dfrac{( x-2) (x-3) }{x-2} =\lim\limits_{x\rightarrow 2^{+}}(x-3) =-1 \end{eqnarray*} \]

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The one-sided limits are not equal, so the derivative does not exist at \(2;\) \(2\) is a critical number.

Step 2 Evaluate \(f\) at the critical numbers \(\dfrac{5}{2}\) and \(2\) and at the endpoints \(0\) and \(3\).

Step 3 The largest value \(3\) is the absolute maximum value; the smallest value \(-1\) is the absolute minimum value.