Question

Find the absolute maximum and absolute minimum values of \(f\) on the given interval.

64. \(f\left( x \right) = x{e^{\frac{x}{2}}},\,\,\,\left( { - 3,1} \right)\)

Short answer

The absolute maximum value of the function \(f\left( x \right) = x{e^{\frac{x}{2}}}\)is \(f\left( 1 \right) \approx 1.649\).

The absolute minimum value of the function \(f\left( x \right) = x{e^{\frac{x}{2}}}\) is \(f\left( { - 2} \right) \approx - 0.736\).

Step-by-step solution

Explanation about Closed Interval Method

To obtain the absolute maximum and lowest values of a continuous function in any closed interval, use these three steps:

• Determine the values of the function at the crucial numbers in while using critical numbers.
• Determine the function's value at each interval's endpoint.
• Determine the highest and smallest values from steps 1 and 2, with both the largest value being absolute maximum and the lowest value being absolute minimum.

Find the critical points by differentiating the given function

The given function is \(f\left( x \right) = x{e^{\frac{x}{2}}}\).

So, differentiating the above function with respect to \(x\) as:

\(\begin{aligned}{c}\frac{d}{{dx}}\left( {f\left( x \right)} \right) &= \frac{d}{{dx}}\left( {x{e^{\frac{x}{2}}}} \right)\\f'\left( x \right) &= {e^{\frac{x}{2}}} \cdot \frac{d}{{dx}}\left( x \right) + x \cdot \frac{d}{{dx}}\left( {{e^{\frac{x}{2}}}} \right)\\ &= {e^{\frac{x}{2}}} + x{e^{\frac{x}{2}}} \cdot \frac{d}{{dx}}\left( {\frac{x}{2}} \right)\\ &= {e^{\frac{x}{2}}} + \frac{{x{e^{\frac{x}{2}}}}}{2}\\ &= \frac{{\left( {x + 2} \right){e^{\frac{x}{2}}}}}{2}\end{aligned}\)

Now, substitute \(f'\left( x \right) = 0\) and solve for \(x\) to get the critical points.

\(\begin{aligned}{c}\frac{{\left( {x + 2} \right){e^{\frac{x}{2}}}}}{2} &= 0\\\left( {x + 2} \right){e^{\frac{x}{2}}} &= 0\\x &= - 2\end{aligned}\)

Find value of \(f\) at critical points in \(\left( { - 3,1} \right)\)

At \(x = - 2\),

\(\begin{aligned}{c}f\left( { - 2} \right) &= \left( { - 2} \right){e^{\frac{{ - 2}}{2}}}\\ &= - 2{e^{ - 1}}\\ &\approx - 0.736\end{aligned}\)

Find value of \(f\) at endpoints of \(\left( { - 3,1} \right)\)

At \(x = - 3\)

\(\begin{aligned}{c}f\left( { - 3} \right) &= \left( { - 3} \right){e^{\frac{{ - 3}}{2}}}\\ &\approx - 0.669\end{aligned}\)

At \(x = 1\)

\(\begin{aligned}{c}f\left( 1 \right) &= \left( 1 \right){e^{\frac{1}{2}}}\\ &\approx 1.649\end{aligned}\)

Now, find the absolute maximum and minimum value from the above steps

By the above two steps maximum and minimum value can be as when \(x = - 2\)the value of \(f\left( { - 2} \right) \approx - 0.736\) that is it is the minimum value and when \(x = 1\)the value of \(f\left( 1 \right) \approx 1.649\) that is it is the maximum value.

Thus, the absolute maximum and minimum values of the function \(f\left( x \right) = x{e^{\frac{x}{2}}}\)are \(f\left( 1 \right) \approx 1.649\) and \(f\left( { - 2} \right) \approx - 0.736\) respectively.