# unique solution

posted by
**Jen**
.

How do I show that the equation

x^4 + 3x + 1 = 0, -2 <= x <= -1

has exactly one solution in the interval.

Thanks.

One way to do this is to use trial and error. split the interval (-2,-1) into 10 equal parts. Then evaluate the function at each point. That is put the value of x as -2, -1.9,-1.8......-1. Look for changes in sign. That is the value of function should change signs, if this happens only once, then the function has only one solution in the interval. The values of x^4 + 3x + 1 at -2, -1.9, -1.8.......-1 are 11, 8.33,6.09, 4.25,2.75,1.56,0.6416,-0.0439,-0.53,-0.84,-1. Its very clear that the sign change occurs only once when X transitions from -1.4 to -1.3 so there is only one solution in this range which is between -1.4 and -1.3. This is however not an exact method.

The exact method would be to show first that the function is monotoniaclly increasing or monotonically decreasing and then show that the function takes a value of 0 in this interval. To do this you need to use calculus.

Yes, calculus is what I want to use!

This is from the chapter "application of derivatives".

Using mean value theorem how can i explain this? Thanks.