# Inverse functions

**Questions**

Find (f^(-1))'(a)

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

a=2

**Answer**

Here is a way.

f^(-1))'(a)

= f^(-1))'(2) because a=2,

= y

That is,

f^(-1))'(2) = y

2 = f(y) by definition of inverse function.

2 = y^5 – y^3 + 2y

y^5 – y^3 + 2y – 2 = 0

y^3[ y^2 – 1 ] + 2[ y – 1 ] = 0 Factorize…

y^3[ (y – 1)*(y + 1) ] + 2[ y – 1 ] = 0

(y -1) [ y^3 (y + 1) + 2 ] = 0

So y – 1 = 0 giving y = 1.OR y^3 (y + 1) + 2 = 0. … (1)

But we evaluate y^3 (y + 1) + 2 for y =0, <0 and > 0.

Consider,

if y= 0, y^3 (y + 1) + 2 = 2 > 0.

if y> 0, y^3 (y + 1) + 2 >0 since all terms are positive,

If y <0 we consider two cases seperately namely

y< -1 and -1 < y <0.

For if y< -1, y^3 <0 and (y+1)<0 and thus y^3(y+1) >0 being the multiplication of tqo negative numbers. and Hence,

y^3 (y + 1) + 2 > 0.

If -1 < y < 0, then 0 < (y + 1) < 1 and -1 < y^3 < 0. and thus,

y^3 (y + 1) is < 0 [ negative ] but its absolute value is less than 1 being the multiplication of two expression whose absolute value of both < 1. That is 0 > y^3 (y + 1) > -1.

Hence, y^3 (y + 1) + 2 > 2-1 =1 > 0.

In other words we have shown that y^3 (y + 1) + 2 is NEVER equal to zero for any “real values” y. This is equivalent to saying that y^3 (y + 1) + 2 = 0 has NO solutions.

Hence y = 1 is the ONLY solution.

Indeed,

f^(-1))'(2) = y = 1.