find the range of these functions algebraically: y =x^3 + 2x^2 - 5x + 2 and...

[hndalama]

a.) y =x³+2x² -5x +2
b.) y=(x²+3x - 5)/(x-4)

for a.), I know that the range is all real numbers because I know the shape the graph will take, however, is their an algebraic way to find the range of a cubic function like this one?

The method I know is to make x the subject of the formula. When I try to do this, I get:
y =x³+2x² -5x +2

My understanding is that because the powers of x are different, it is not possible to make x the subject of the formula. Is there an algebraic way of determining the range?

b.) y=(x²+3x - 5)/(x-4)
for this one I get the same problem as well.
y(x-4) = x²+3x - 5

I again have different powers of x so it doesn't seem possible to find the range using this method. Is there an algebraic way of determining the range?

 

[stapel]

Find the range of these functions algebraically:

a) y = x³ + 2x² - 5x + 2

For (a), I know that the range is all real numbers because I know the shape the graph will take. However, is there an algebraic way to find the range of a cubic function like this one?

The method I know is to make x the subject of the formula. When I try to do this, I get:
y =x³+2x² -5x +2

How is this different from what you'd started with?

My understanding is that because the powers of x are different, it is not possible to make x the subject of the formula. Is there an algebraic way of determining the range?

Usually, this is done with a graph, but the exercise specifies to use algebra, rather than a picture. How did you textbook and instructor do similar exercises?

Off the top of my head, you could convert this to a (trivial) rational function:

. . . . .\(\displaystyle y\, =\, \dfrac{x^4\, +\, 2x^3\, -\, 5x^2\, +\, 2x}{x}\)

Then apply the algebraic rules for horizontal asymptotes.

b) y = (x² + 3x - 5) / (x - 4)
for this one I get the same problem as well.

This one is already set up for the horizontal asymptotes rules.

 

[hndalama]

How is this different from what you'd started with?


Usually, this is done with a graph, but the exercise specifies to use algebra, rather than a picture. How did you textbook and instructor do similar exercises?

The method I've been taught is to make x the subject of the formula and then find the domain of the equation that results. However, in both equations it's not possible to make x the subject of the formula so that's why I am stuck on the first step.

apply the algebraic rules for horizontal asymptotes.

The algebraic method I know for finding the horizontal asymptote is to take the limit to +or - infinity. For equation a, the limit is plus or minus infinity, hence this gives the correct range of all real values.

Taking the limit to plus or minus infinity of equation b, gives:

limit to infinity of: (x²+3x - 5)/(x-4) = limit to infinity of: (1 + 3/x - 5/x²)/(1/x -4/x²) = plus or minus infinity.

So this also gives the range as all real values but I graphed the equation on desmos and found that the range is y<=1.408 .

Can this range be found algebraically?

 

[stapel]

The algebraic method I know for finding the horizontal asymptote is to take the limit to +or - infinity. For equation (a), the limit is plus or minus infinity, hence this gives the correct range of all real values.

Taking the limit to plus or minus infinity of equation b, gives:

limit to infinity of: (x²+3x - 5)/(x-4) = limit to infinity of: (1 + 3/x - 5/x²)/(1/x -4/x²) = plus or minus infinity.

So this also gives the range as all real values but I graphed the equation on desmos and found that the range is y<=1.408 .

I have no idea what you might have entered into "desmos", but one can easily show that this range is not correct. Plug in x = 5:

. . . . .\(\displaystyle \dfrac{(5)^2\, +\, 3(5)\, -\, 5}{(5)\, -\, 4}\, =\, \dfrac{25\, +\, 15\, -\, 5}{1}\, =\, 35\)

This is certainly greater than 1.408. Did you perhaps use a small graphing window, so you took only the max of the function to the left of the vertical asymptote?

(This highlights how the algebra, such as the rules for horizontal asymptotes, can be more reliable than pictures.)

 

[mmm4444bot]

a.) y = x³ + 2x² - 5x + 2

I know that the range is all real numbers because I know the shape the graph will take, however, is [there] an algebraic way to find the range of a cubic function like this one?

You could take the limit of y, as x increases (or decreases) without bound. That is, investigate the behavior of y, when x approaches positive infinity and when x approaches negative infinity.

With polynomials, it's the leading term (i.e., the highest-degree term) that controls these behaviors. In other words, the behavior of x^3+2x^2-5x+2 is the same as the behavior of x^3, when the absolute value of x is insanely large.

If the limit concept is unfamiliar to you, then just use your knowledge of the shape of a cubic polynomial's graph (like you did), or memorize the fact that the domain and range for all polynomial functions with odd degree are "all Real numbers".

The method I know is to make x the subject of the formula. … My understanding is that because the powers of x are different, it is not possible to make x the subject of the formula.

Are you confusing the process with finding an inverse function? (We could find piecewise inverse functions, by restricting the domain.)

To make x the subject of the given equation (i.e., to solve for x), we could apply the Cubic Formula.

Don't do either of these things; there's an easier way.

b.) y = (x² + 3x - 5)/(x - 4)

for this one I get the same problem as well.

y(x - 4) = x² + 3x - 5

I again have different powers of x so it doesn't seem possible to find the range using this method. Is there an algebraic way of determining the range?

You've made the first step, in a different method!

If you multiply out the left-hand side, and then subtract every term on the right-hand side from each side, you will get a quadratic equation with some symbolic coefficients (i.e., B and C will be expressed in terms of y).

A*x^2 + B*x + C = 0

We would use the Quadratic Formula, were we to want to solve for x, but we don't need to do that. We just need to realize that, in the Quadratic Formula, the Discriminant must be greater than or equal to zero, in order for x to be Real.

Therefore, once you have your coefficients A, B, and C, solve the following inequality for y.

B^2 - 4AC ≥ 0

You will get two Real solutions for y. Use those solutions, to write the range of the Rational Function in part b.

If you need help visualizing which solution goes where, look at a graph of the function or analyze some evaluated test values.

Questions? Please show us what you tried. :cool:

 

[hndalama]

This is certainly greater than 1.408. Did you perhaps use a small graphing window, so you took only the max of the function to the left of the vertical asymptote?

Yes, I didn't see the whole graph. The range is actually 1.408>= y >= 20.59.

use the Quadratic Formula

I was able to find the same range by using the quadratic formula. This seems to be the way to find the range algebraically.

Thank you

 

[tkhunny]

b.) y=(x²+3x - 5)/(x-4)

The method I know is to make x the subject of the formula.

Why? What is that supposed to do?

It's a Rational Function. Does it have a vertical asymptote? Is the factor that caused the vertical asymptote an odd power? Does it have an Oblique Asymptote? Have you studied Rational Functions?

Rule of Thumb: Start with All Real Numbers and see if you can rule anything out.

 

[hndalama]

Why? What is that supposed to do?

The domain of the inverse equals the range of the function

 

[mmm4444bot]

… I was able to find the same range by using the quadratic formula.

What quadratic equation did you use, for this?

 

[tkhunny]

The domain of the inverse equals the range of the function

And how does this help? You still have to track it down in just the same way.

 

[hndalama]

What quadratic equation did you use, for this?

y=(x²+3x - 5)/(x-4)
y(x-4) = x²+3x-5
yx-4y =x²+3x-5
x²+(3-y)x +4y-5 = 0

so x= [-(3-y)+/- √[(3-y)² -4(4y-5)] ] / 2
x= [-(3-y)+/- √[(3-y)² -4(4y-5)] ] / 2[FONT=&quot]
The domain of this equation is all the values ofy where the expression inside the square root is not negative. that is allvalues of y where y[/FONT]^{[FONT=&quot]2[/FONT]}[FONT=&quot]-22y+29>=0

solve the roots of the expression using the quadratic formula. These are thecritical points which can then be used to determine where the expression is nonnegative.
This domain is the range of the original function[/FONT]

 

[mmm4444bot]

y = (x^2 + 3x - 5) / (x - 4)


… x^2 + (3 - y)x + (4y - 5) = 0

… x = [-(3 - y) +/- √[(3 - y)2 - 4(4y - 5)]] / 2

The domain of this equation is all the values of y where the expression inside the square root is not negative. that is all values of y where

y^2 - 22y + 29 >= 0


… This domain is the range of the original function

… solve the roots of the expression …

… to determine where the expression is nonnegative …

This is the method through which I guided you earlier.

I wanted to see your Quadratic Equation because you misstated the range, in your prior post:

The range is actually 1.408 >= y >= 20.59

Those values are correct (although, I'm not sure why you've rounded one to the nearest thousandth of a unit, while rounding the other to the nearest hundredth of a unit), yet, the compound inequality above is not a correct statement for the range.

You have actually inferred that y is between 1.408 and 20.59, but it's not. Additionally, by using ≥ symbols with the values shown in that order, you've stated that 1.408 is a larger number than 20.59!

Do you see these issues? That is, can you state the inequalities (that is, the range) properly?

PS: The name for the expression inside a square root symbol is "radicand". We're talking about a specific radicand (the one that appears in the Quadratic Formula). It's special enough to have its own name, so, instead of referring to B^2-4AC as "the expression inside the square root", you could call it by name: the Discriminant. :cool: