Calculus- Early Transcendentals, 2021a

112 Limits
f (x) is continuous on [0,8]. Wehavef (a)= f (0)= 3√ 0+0−1 = −1and f (b)= f (8)= 3√ 8+8−1 = 9.
Thus N = 0 lies between f (a)=−1and f (b)=9, so the conditions of the Intermediate Value Theorem are
satisfied. So, there exists a number c in (0,8) such that f (c)=0. This means that c satisfies 3√ c+c−1 = 0,
in otherwords, is a solution for the equation given.
Alternatively we can let f (x)= 3√ x +x, N = 1, a = 0andb = 8. Then as before f (x) is the sum of two
continuous functions, so is also continuous everywhere, in particular, continuous on the interval [0,8]. We
have f (a)= f (0)= 3√ 0 + 0 = 0and f (b)= f (8)= 3√ 8 + 8 = 10. Thus N = 1 lies between f (a)=0and
f (b)=10, so the conditions of the Intermediate Value Theorem are satisfied. So, there exists a number
c in (0,8) such that f (c)=1. This means that c satisfies 3√ c + c = 1, in otherwords, is a solution for the
equation given.
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Example 3.56: Roots of Function
Explain why the function f = x 3 + 3x 2 + x − 2 has a root between 0 and 1.
Solution. By Theorem 3.8, f is continuous. Since f (0)=−2 and f (1)=3, and 0 is between −2 and 3,
there is a c ∈ (0,1) such that f (c)=0.
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This example also points the way to a simple method for approximating roots.
Example 3.57: Approximating Roots
Approximate the root of the previous example to one decimal place.
Solution. If we compute f (0.1), f (0.2), and so on, we find that f (0.6) < 0and f (0.7) > 0, so by the
Intermediate Value Theorem, f has a root between 0.6 and 0.7. Repeating the process with f (0.61),
f (0.62), and so on, we find that f (0.61) < 0and f (0.62) > 0, so f has a root between 0.61 and 0.62, and
the root is 0.6 rounded to one decimal place.
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Exercises for 3.7
Exercise 3.7.1 Consider the function
h(x)=
{
2x − 3, if x < 1
0, if x ≥ 1
Show that it is continuous at the point x = 0. Is h a continuous function?
Exercise 3.7.2 Find the values of a that make the function f (x) continuous for all real numbers.
{ 4x + 5, if x ≥−2
f (x)=
x 2 + a, if x < −2