3. Absolute Extrema. Use the derivative and limits to find absolute (global) maximum and minimum values, if they exist.

Question

Absolute Extrema.  Use the derivative and limits to find absolute (global) maximum and minimum values, if they exist.

f(x) = 6e^x - x^2 / e^x + 2x^2

Answer 1

When a question asks for absolute extrema on a function like this one, the first thing worth noticing is that the denominator e^x + 2x² is never zero, since e^x is always positive and 2x² is never negative. So f is defined for every real number and we are looking across the whole line, not a closed interval. Reading the function as

f(x) = (6e^x − x²) / (e^x + 2x²)

there is a shortcut here that is much quicker than the derivative. If we suspect a number is the maximum, we can subtract it and look at the sign of whatever is left. So, let's try 6:

f(x) − 6 = (6e^x − x² − 6e^x − 12x²) / (e^x + 2x²) = −13x² / (e^x + 2x²)

The top is never positive and the bottom is always positive, so f(x) − 6 is never positive. That tells us f(x) is less than or equal to 6 everywhere, and the top is only zero when x = 0. Since f(0) = 6/1 = 6, the absolute maximum value is 6 and it happens at x = 0.

Now for the minimum, we can do the same thing with −1/2:

f(x) + 1/2 = (6e^x − x² + (1/2)e^x + x²) / (e^x + 2x²) = (13/2)e^x / (e^x + 2x²)

Since e^x is always positive, this whole thing is always positive, which means f(x) is always greater than −1/2 and never equal to it. The limits show where that number comes from. As x goes to negative infinity, e^x goes to 0, so the function behaves like −x² / 2x², which is −1/2. As x goes to positive infinity the e^x terms take over and it behaves like 6e^x / e^x, which is 6.

So the graph creeps down toward −1/2 on the far left and flattens toward 6 on the far right, touching 6 exactly once at x = 0. Remember that an absolute minimum has to be a value the function actually reaches, and −1/2 never is. So, our final answer is an absolute maximum of 6 at x = 0, and no absolute minimum.