SOLUTION: Given f(x) = x^4-2x^2-15x 1. Find the x-intercepts of the function f(x) I Have This x^3-2x^2-15x=0 x(x+3)(x-5) x+3=0 x-5=0 x=0 x=-3 x=5 (-3,0), (5,0),

Question 181524This question is from textbook College Algebra the Graphing Approach
: Given f(x) = x^4-2x^2-15x
1. Find the x-intercepts of the function f(x)
I Have This
x^3-2x^2-15x=0
x(x+3)(x-5)
x+3=0 x-5=0 x=0
x=-3 x=5
(-3,0), (5,0), (0,0)
I'm having trouble with #3 and 4
3. What is the domain and range of f(x)
4. Over what ranges is f(x) and increasing and decreasing function.
These problems work off the f(x) = x^3-2x^2-15x
Hope you can help me. Thanks This question is from textbook College Algebra the Graphing Approach

Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

You have correctly identified the zeroes of this function.

The domain of a function is that set of values for which the function is defined. In some cases, there are values that would make a function undefined, such as a value that would cause a denominator to be zero or a radicand to be less than zero. In the case of all polynomial functions, such as your very handsome little cubic, there are no real numbers that would make the function undefined. Therefore the domain is all real numbers; in set builder notation:

The range of a function is that set of values of the function for every value of the independent variable in the range. Polynomial functions of odd degree (yours is degree 3) have the characteristic of having a range that is all real numbers.

This makes sense intuitively because if you consider a very large negative number for x, raising that number to the 3rd power will result in an extremely large magnitude negative number -- in other words, as x decreases without bound, so does f(x). Likewise, as as x increases without bound, so does f(x).

If you graph this, you will see that the curve comes up, crosses the x-axis at (-3, 0) and increases to a point somewhere between (-3, 0) and (0, 0). It then starts to decrease, crossing the x-axis again at (0, 0) to a point somewhere between (0, 0) and (5, 0).

You can find these points by the process of taking the first derivitive of the function and setting it equal to zero if the derivitive exists. The point is that the first derivitive is the instantaneous rate of change of the function at x, so if you find points where the rate of change is zero, you have a local maximum or a local minimum. If the second derivitive exists, then the sign on the value of the second derivitive at the minimum or maximum tells you whether the point is a minimum or a maximum ( minus means a maximum, plus means a minimum.

I'll let you do your own arithmetic to verify this, but one of the points is:

and the other is:

Now:

Negative, therefore

is a maximum.

Positive, therefore

is a minimum.

Having found these points we can now define the intervals in which the function increases and decreases. Since

is a maximum, the function must increase to get there, so it is increasing on the interval

. It decreases on the interval

. And finally, it increases on the interval

. Notice that none of the endpoints of the intervals are included. Infinity never has the endpoint included because there isn't one, and the other points represent points where the function is neither increasing nor decreasing.

John

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