Finding the range of a $y=-x^2(x+5)(x-3)$ without calculus?

Question

I was helping a precalculus student with this question. The graph wasn't given. My only idea was to find the inverse and try to find its domain. When trying to find the inverse, I arrived at $0=y^4+2y^3-15y^2+x$ and I was hoping to be able to factor it. However, besides using the quartic formula, I'm not sure how to do this.

Answer 1

Definitely don't need calculus nor a graphing calculator for this problem. We are trying to find $$-x^4-2x^3+15x^2=y_{max}$$ Rewrite as $$x^4+2x^3-15x^2+y_{max}=0$$ Now if we knew where the maximum occurred, say at $x=a$, then $a$ would be a root of this equation. Also it must be a double root of else the graph of the function would cross the $x$-axis at this point, so it wouldn't be a maximum. Thus $(x-a)^2=x^2-2ax+a^2$ must be a factor of $x^4+2x^3-15x^2+y_{max}$. Carrying out polynomial long division, we get a quotient of $$x^2+(2a+2)x+(3a^2+4a-15)$$ and a remainder of $$(4a^3+6a^2-30a)x+y_{max}-(3a^4+4a^3-15a^2)$$ The quotient is of little interest to us, but the remainder must be the zero polynomial, so $$4a^3+6a^2-30a=2a(2a^2+3a-15)=0$$ Here we can recognize the derivative, but no calculus was harmed in the production of this post! So $$a\in\left\{0,\frac{-3+\sqrt{129}}4,\frac{-3-\sqrt{129}}4\right\}$$ And since $$y_{max}=3a^4+4a^3-15a^2$$ We can just test the $3$ roots and get values of $0$, $28.182551$, and $119.754949$ respectively. Thus the max occurs at $a=\frac{-3-\sqrt{129}}4\approx-3.598454$ and has the value $y_{max}=3a^4+4a^3-15a^2\approx119.754949$.

EDIT: I thought that it would be tedious be expand that polynomial in $a$ to find the exact value of $y_{max}$, but if we observe that $y_{max}=\frac{u+v\sqrt{129}}{256}\approx119.754949$ and then $\frac{u-v\sqrt{129}}{256}\approx28.182551$, where $u$ and $v$ are integers, we see that $u=128(119.754949+28.182551)=18936=8\cdot2367$ and $v=128(119.754949-28.182551)/\sqrt{129}=1032=8\cdot129$, we only have to be sure of our approximate results for $u$ and $v$ to the nearest integer to obtain the exact result $y_{max}=\frac{2367+129\sqrt{129}}{32}$.

Answer 2

Well, the function is always negative if $x<-5$ or $x>3$, is equal to zero when $x$ is $-5$ or $3$, and grows in magnitude without bound as $x$ increases past $3$, so the range includes $(-∞,0]$.

Between $-5$ and $3$ the function is positive (except at $0$), so it's a question of finding the maximum achieved in this range. Personally, if I wasn't using calculus, I would graph the function with a graphing calculator, observe that the maximum is about $119.75$, and conclude that the range is approximately $(-\infty, 119.75]$.

Answer 3

First we have to find the zeros of the eqn $y=-(x^2)(x+5)(x-3)$. So from the given eqn we get the zeros which are 0 of multiplicity 2,-5 of 1 and 3 of 1.

1.$y>0$ when $-5<x<3$

2.$y<0$ when $-\infty<x<-5\;\;and\;\;3<x<\infty$

3.As $x$ tends to $-\infty$, $y$ tends to $-\infty$ and $x$ tends to +infinty, $y$ tends to -infinity.

Now we have to calculate the max value of $y$.

$y=-(x^2)(x+5)(x-3)$

$\implies$$y=-(x^2)({x^2}+2x-15)$

$\implies$$y=-({x^4}+2{x^3}-15{x^2})$

Then make this term ${x^4}+2{x^3}-15{x^2}$ into of the form

$y=-({f(x)^4})+K_1\;\;where\;\;K_1>0$

Or

$y=-({f(x)^4}+{g(x)^2})+K_2\text{ where }K_2>0$

Then you will get $K_1=K_2$, the max value of $y$.

Then you will get the $R(y)=(-\infty,K_1\;\;OR\;\;K_2$]