Tuesday

July 22, 2014

July 22, 2014

Posted by **Jen** on Sunday, November 5, 2006 at 8:55pm.

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.

**Related Questions**

algebra - Which has a Unique solution, No solution, or infinitely many ...

Calculas HS - show that the solution has exactly one solution in the interval x+...

math - two systems of equations are giving below for each system choosethe best ...

Calculus - True or false: If f(x) is continuous on the interval [a,b] & if the ...

calc - analyze the function ln x = cx^2 to find the unique value of c such that ...

Calculus AB - Analyze the function ln x=cx^2 to find the unique value of c such...

Advanced Functions - Consider the trig equation: 4sinxcos2x + 4cosxsin2x - 1 = 0...

CountIblis, can i bother you - Countiblis, i know that you might be busy do you ...

Algebra - Two more Algebra questions : Drawing a blank on this one, not sure ...

Algebra - Two more Algebra questions : Having some difficulty with these. ...