SOLUTION: Find the Inverse of {{{ f(x)=3+x^3 }}} Use Descarte's Rule of Signes to determine the number of positive and negative real zeros of the polynomial {{{ p(x) = 2x^4+3x^3-2x^2+x-2

Algebra ->  Inequalities -> SOLUTION: Find the Inverse of {{{ f(x)=3+x^3 }}} Use Descarte's Rule of Signes to determine the number of positive and negative real zeros of the polynomial {{{ p(x) = 2x^4+3x^3-2x^2+x-2       Log On


   

Question 153021: Find the Inverse of +f%28x%29=3%2Bx%5E3+
Use Descarte's Rule of Signes to determine the number of positive and negative real zeros of the polynomial +p%28x%29+=+2x%5E4%2B3x%5E3-2x%5E2%2Bx-2+ can have

Found 2 solutions by jim_thompson5910, Earlsdon:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
# 1)

y=3%2Bx%5E3 Start with the given equation.


x=3%2By%5E3 Switch x and y


x-3=y%5E3 Subtract 3 from both sides.


root%283%2Cx-3%29=y Subtract 3 from both sides.


y=root%283%2Cx-3%29


So the inverse function is




# 2)

Positive Zeros:



First count the sign changes of f%28x%29=2x%5E4%2B3x%5E3-2x%5E2%2Bx-2

From 2x%5E4 to 3x%5E3, there is no change in sign

From 3x%5E3 to -2x%5E2, there is a sign change from positive to negative

From -2x%5E2 to x, there is a sign change from negative to positive

From x to -2, there is a sign change from positive to negative

So there are 3 sign changes for the expression f%28x%29=2x%5E4%2B3x%5E3-2x%5E2%2Bx-2.

So there are 3 or 1 positive zeros



------------------------------------------------


Negative Zeros:



f%28-x%29=2%28-x%29%5E4%2B3%28-x%29%5E3-2%28-x%29%5E2%2B%28-x%29-2 Now let's replace each x with -x


f%28-x%29=2x%5E4-3x%5E3-2x%5E2-x-2 Simplify


Now let's count the sign changes of f%28-x%29=2x%5E4-3x%5E3-2x%5E2-x-2

From 2x%5E4 to -3x%5E3, there is a sign change from positive to negative

From -3x%5E3 to -2x%5E2, there is no change in sign

From -2x%5E2 to -x, there is no change in sign

From -x to -2, there is no change in sign

So there is 1 sign change for the expression f%28-x%29=2x%5E4-3x%5E3-2x%5E2-x-2.

So there is 1 negative zero

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
1) Find the inverse of:
f%28x%29+=+3%2Bx%5E3 Rewrite this as:
y+=+3%2Bx%5E3 and exchange the x and y to get:
x+=+3%2By%5E3 Now solve this for y.
x-3+=+y%5E3 Take the cube root of both sides.
y+=+root%283%2C%28x-3%29%29
Now replace the y with f%5E%28-1%29%28x%29 to get:
f%5E%28-1%29%28x%29+=+root%283%2C%28x-3%29%29 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%28x%29+=+2x%5E4%2B3x%5E3-2x%5E2%2Bx-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%28-x%29+=+2%28-x%29%5E4%2B3%28-x%29%5E3-2%28-x%29%5E2%2B%28-x%29-2 Making the appropriate sign changes:
p%28-x%29+=+2%28x%29%5E4-3%28x%29%5E3-2%28x%29%5E2-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.