Power Functions can be defined as functions which are written in form of y = xn. So according to definition of power function we can say that Polynomial Functions consist of power functions. For example: Volume of any surface rises as (4/2) th power of surface area. Now we will see how to solve power functions. Before solving power functions we need to understand rules of exponents or powers.
· x (u + v) = (xu) (xv)
· x u v = (xu)v
· x -u = 1 / xu
· x (1/u) is the u-th root of x.
· x0 = 1 for any x ≠ 0. And we know that the power of 0 is undefined.
· (xy) u = (xu) (xv)
Now we will see how to solve power functions. We need to follow some steps.
Step 1: First we take a function which contains power values. Suppose we have a function i.e. f (p) = p6. Find value of f (2) and find 'p' such that f (p) = 80.
Step 2: If we put value of 'p' as 2 in given function we get:
= f (p) = p6, now put p = 2,
= f (2) = 26, on further solving we get,
= f (2) = 2* 2* 2* 2* 2* 2 = 64.
For second question: f (p) = 80.
Step 3: Function can also be written as:
=> p6 = 80.
Step 4: If we write this function in such a way that it does not affect original value. So it can be written as:
= (p6)1/6 = + 80 (1/6).
So this is equals to + 2.075. In this way we solve power functions.