Calculus- Early Transcendentals, 2021a

110 Limits
Example 3.53: Motivation for the Intermediate Value Theorem
1. Does e x + x 2 = 0 have a solution?
2. Does e x + x = 0 have a solution?
Solution.
1. The first question is easy to answer since for any exponential function we know that a x > 0, and
we also know that whenever you square a number you get a nonnegative answer: x 2 ≥ 0. Hence,
e x + x 2 > 0, and thus, is never equal to zero. Therefore, the first equation has no solution.
2. For the second question, it is difficult to see if e x + x = 0 has a solution. If we tried to solve for x,
we would run into problems. Let’s make a table of values to see what kind of values we get (recall
that e ≈ 2.7183):
x e x + x
−2 e −2 − 2 ≈−1.9
−1 e −1 − 1 ≈−0.6
0 e 0 + 0 = 1
1 e + 1 ≈ 3.7
Sketching this gives:
Let f (x)=e x + x. Notice that if we choose a = −1 andb = 0thenwehave f (a) < 0and f (b) > 0.
A point where the function f (x) crosses the x-axis gives a solution to e x +x = 0. Since f (x)=e x +x
is continuous (both e x and x are continuous), then the function must cross the x-axis somewhere
between −1 and 0: