Calculus- Early Transcendentals, 2021a

3.7. Continuity 111
Therefore, our equation has a solution.
Note that by looking at smaller and smaller intervals (a,b) with f (a) < 0and f (b) > 0, we can
get a better and better approximation for a solution to e x + x = 0. For example, taking the interval
(−0.4,−0.6) gives f (−0.4) < 0and f (−0.6) > 0, thus, there is a solution to f (x)=0 between −0.4
and −0.6. It turns out that the solution to e x + x = 0isx ≈−0.56714.
We now generalize the argument used in the previous example. In that example we had a continuous
function that went from negative to positive and hence, had to cross the x-axis at some point. In fact, we
don’t need to use the x-axis, any line y = N will work so long as the function is continuous and below the
line y = N at some point and above the line y = N at another point. This is known as the Intermediate
Values Theorem and it is formally stated as follows:
Theorem 3.54: Intermediate Value Theorem
If f is continuous on the interval [a,b] and N is between f (a) and f (b), where f (a) ≠ f (b), then
there is a number c in (a,b) such that f (c)=N.
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The Intermediate Value Theorem guarantees that if f (x) is continuous and f (a) < N < f (b), the line
y = N intersects the function at some point x = c. Such a number c is between a and b and has the property
that f (c)=N (see Figure 3.5(a)). We can also think of the theorem as saying if we draw the line y = N
between the lines y = f (a) and y = f (b), then the function cannot jump over the line y = N. On the other
hand, if f (x) is not continuous, then the theorem may not hold. See Figure 3.5(b) where there is no number
c in (a,b) such that f (c) =N. Finally, we remark that there may be multiple choices for c (i.e., lots of
numbers between a and b with y-coordinate N). See Figure 3.5(c) for such an example.
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Figure 3.5: (a) A continuous function where IVT holds for a single value c. (b) A discontinuous
function where IVT fails to hold. (c) A continuous function where IVT holds for multiple
values in (a,b).
The Intermediate Value Theorem is most frequently used for N = 0.
Example 3.55: Intermediate Value Theorem
Show that there is a solution of 3√ x + x = 1 in the interval (0,8).
Solution. Let f (x)= 3√ x+x−1, N = 0, a = 0, and b = 8. Since 3√ x, x and −1 are continuous on R,andthe
sum of continuous functions is again continuous, we have that f (x) is continuous on R, thus in particular,