# How do you use the intermediate value theorem to prove that #f(x)=x^3-9x+5# has a real zero in the intervals [-4,-3], [0,1], and [2,3]?

##### 1 Answer

See explanation...

#### Explanation:

Given:

#f(x) = x^3-9x+5#

By the intermediate value theorem (or by the simpler Bolzano's theorem), the function

With the given

#{ (f(-4) = (color(blue)(-4))^3-9(color(blue)(-4))+5 = -64+36+5 = -23 < 0), (f(-3) = (color(blue)(-3))^3-9(color(blue)(-3))+5 = -27+27+5 = 5 > 0) :}#

So

#{ (f(0) = (color(blue)(0))^3-9(color(blue)(0))+5 = 5 > 0), (f(1) = (color(blue)(1))^3-9(color(blue)(1))+5 = 1-9+5 = -3 < 0) :}#

So

#{ (f(2) = (color(blue)(2))^3-9(color(blue)(2))+5 = 8-18+5 = -5 < 0), (f(3) = (color(blue)(3))^3-9(color(blue)(3))+5 = 27-27+5 = 5 > 0) :}#

So

graph{x^3-9x+5 [-7, 7, -10, 20]}

**Notes**

The intermediate value theorem is one way of expressing the *completeness* of the real numbers. It basically says that there are no gaps in the real number line that would allow a continuous function to change sign without being equal to zero at some point.

The same is not true of the rational numbers. For example, the function:

#f(x) = x^2-2#

has the properties:

#{ (f(1) = -1 < 0), (f(2) = 2 > 0) :}#

but there is no rational number

So there are "gaps" in the rational numbers - in this particular case the number we would call