Expert: Paul Klarreich - 10/16/2007

Question
Hi
I needed help with the following problem:
Given that f(x)=x^4-x^2+5x+2, show that there exist at least two real numbers x so f(x)=3. How do i show this? Thanks

Answer
Questioner:   Dragos
Category:  Calculus
Private:  No
 
Subject:  Calculus I
Question:  Hi
I needed help with the following problem:
Given that f(x)=x^4-x^2+5x+2, show that there exist at least two real numbers x so f(x)=3. How do i show this? Thanks
................................
Hi, Dragos,

Thank you for the kind comments.  You are, however, correct in that I was not at all clear about one item.  (That's what comes when I overuse the 'cut and paste' operations.  Serves me right.)

You will find the new language below, between the ********* marks.
------------ ORIGINAL ASNWER -------------
This looks like an Intermediate Value example.

Principle:  if f(x1) = y1   and  f(x2) = y2,  then  f(x) takes on every value between y1 and y2  somewhere  between  x1 and x2.  

In words, if f(x) takes on two values, it must take on all the values between them.

So if we find

f(x1) that is less than 3, and
f(x2) that is more than 3, and
f(x) is continuous, then somewhere between x1 and x2, f(x) must be equal to 3.

Now the example:

f(x) = x^4 - x^2 + 5x + 2

This is a polynomial of degree 4, which is even, and has a leading coefficient of +1.

So it is certainly true that as  x -> PLUS infinity, so does f(x), AND  f(0) = 2, so there must be a value of x  in (2, +inf) where f(x) = 3.
[there's one value]
************* MODIFIED *****************
Further it is certainly true that as x --> MINUS infinity, f(x) approaches PLUS INFINITY, since this is a polynomial of even degree, AND since f(0) = 2, there must be a value of x  in (-inf, 2) where f(x) = 3.
[and there's your second]
***************************************
Note that you were never asked to find these numbers, only to show that they exist.

Looking at the graph, you are saying:

On the left side, the graph is very high, certainly above 3, but it comes down to f(0) = 2, which is below 3.  So it must have hit 3, at some negative x.

On the right side, the graph starts at f(0) = 2, which is below 3, but it gets very high, certainly above 3.  So it must hit 3, at some positive x.