inverse functions given a real number

Calculus
Tutor: None Selected Time limit: 1 Day

f(x) =2x^5+x^3+1 a=-2

Apr 22nd, 2015

We need to find f(-2)

Needed is the value of x which solves the equation

2x^5+x^3+1= - 2

or

2x^5+x^3+3=0

which we can write as:

2x^5+2+x^3+1=0

Using polynomial division it is easy to show that:

x^5+1= (x+1) ( x^4-x^3+x^2-x+1)

x^3+1= (x+1) (x^2-x+1) ( in fact there is similar  formula for any odd n for x^n+1)

Thus (x+1) factors out of the equation and we can set

x+1 =0

and one of the possible solutions is

x=-1

There could be 4 more solutions (in this case they are complex) but we'll not work them out here.

Thus f(-2)=-1

Apr 22nd, 2015

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