Question 153021
1) Find the inverse of:
{{{f(x) = 3+x^3}}} Rewrite this as:
{{{y = 3+x^3}}} and exchange the x and y to get:
{{{x = 3+y^3}}} Now solve this for y.
{{{x-3 = y^3}}} Take the cube root of both sides.
{{{y = root(3,(x-3))}}}
Now replace the y with {{{f^(-1)(x)}}} to get:
{{{f^(-1)(x) = root(3,(x-3))}}} This is the inverse of the given function.

2) Use Descartes' rule of signs to find the number of  positive and negative real zeros.
{{{p(x) = 2x^4+3x^3-2x^2+x-2}}} First, for the positive zeros, count the number of sign changes in the polynomial p(x):
There are three sign changes, so, there is a maximum of three real positive zeros, but (counting down by two's) there could also be only one real positive zero.
Let's look at the possible negative zeros by evaluating the function at (-x):
{{{p(-x) = 2(-x)^4+3(-x)^3-2(-x)^2+(-x)-2}}} Making the appropriate sign changes:
{{{p(-x) = 2(x)^4-3(x)^3-2(x)^2-x-2}}} Here, we see only one sign change, so you can expect only one negative root real zero.
In summary, the number of zeros (real or complex) for this function should be 4, because you have a fourth-order polynomial.  However, some of those zeros may be complex.
In fact, if you can work it out, the function has two real zeros, one positive and one negative, and it has two complex zeros.