Question

lets try that again
can anyone find the range of \[f(x)=x^2-\sqrt{x}\] WITHOUT using calculus?

Answer 1

can anyone find the range of \[f(x)=x^2-\sqrt{x}\] WITHOUT using calculus?

Answer 2

i think the domain will be only +ve real numbers ..and zero
so ... range should be from zero

[0,infi ) i'm not sure but thats what i can make out right now

Answer 3

\[x=\frac{1}{2}\]
\[f(\frac{1}{2})=\frac{1}{4}-\frac{\sqrt{2}}{2}<0\]

Answer 4

I suppose u could just test 0,1 0,-1 and + -

Answer 5

i really appreciate your help, satellite73!

Answer 6

well i don't know how to do this without calc and i don't know what your teacher expects you to do. but the answer you had is wrong.

Answer 7

satelitte, I have a question, is range from -0.47247 to infinity ?

Answer 8

the x^2 takes control near infinity so id say its [0,inf) as well

Answer 9

that came out wrong lol

Answer 10

ideally id say when sqrt(x) is twice as large as x^2 we could test that point....

sqrt(x) = 2x^2
x = 4x^4
x/x^4 = 4

1/x^3 = 4

x^3 = 1/4
x = cbrt(1/4) .... maybe?

Answer 11

i spose twice as large superfluous; 5times as large etc amount to the same logic doesnt it

Answer 12

amistre, check x=0.47247

Answer 13

I can't find a way to solve it without any calculus, but I can do it using basic notions of limits and extrema, without getting too rigorous or using derivatives:

The domain is clearly \[[0, \infty).\]\[f(0) = 0,\ f\left(\frac{1}{2}\right) = \frac{1}{4} - \frac{\sqrt{2}}{2}<0, \ f(4) = 14 > 0\]so there is some interval where the function is decreasing and some interval where the function is increasing. We know that \[x^2 > \sqrt{x}\]if \[x > 1\]and\[x^2 < \sqrt{x}\]if\[0 < x < 1\]so the range of the function is \[(0,\infty)\] on the first interval,\[(1, \infty).\]We know it takes negative values in the second interval and our goal from now on will be to determine how negative those values are. Since\[f(1) = 0,\]the function must start increasing somewhere in\[(0, 1).\]The point where the function starts increasing again will be the minimum (since it's decreasing until that point and then it starts increasing). Let \[1 < \varepsilon < 0\]and\[x < 1\]Suppose \[\varepsilon\]is small enough so \[x + \varepsilon < 1.\]If \[x\]is in the decreasing part of the interval and if \[\varepsilon\]is small enough we must have\[x^2 - \sqrt{x} > (x + \varepsilon)^2 - \sqrt{x + \varepsilon}\](which is the definition of a decreasing function). Squaring what must be squared and tidying up a bit we get\[\varepsilon^2\left(1 - 16 x^3 - 24 x^2 \varepsilon - 12 x \varepsilon^2 +
16 x^4 \varepsilon^2 - 2 \varepsilon^3 + 32 x^3 \varepsilon^3 +
24 x^2 \varepsilon^4 + 8 x \varepsilon^5 + \varepsilon^6\right) > 0.\]Because \[\varepsilon > 0,\]we get\[1 - 16 x^3 + \varepsilon \left(-24 x^2 - 12 x \varepsilon + 16 x^4 \varepsilon -
2 \varepsilon^2 + 32 x^3 \varepsilon^2 + 24 x^2 \varepsilon^3 +
8 x \varepsilon^4 + \varepsilon^5\right) > 0.\]Now, we said\[\varepsilon\]is small. ¿How small? Well, as small as we want it to be. Because x is constant (we aren't changing its value) \[-24 x^2 - 12 x \varepsilon + 16 x^4 \varepsilon -
2 \varepsilon^2 + 32 x^3 \varepsilon^2 + 24 x^2 \varepsilon^3 +
8 x \varepsilon^4 + \varepsilon^5\]is some number that doesn't get too big. Because it's multiplied by \[\varepsilon\]which is a small number, the product is small too (it may be positive or negative, but its absolute value is small). Since this product is arbitrarily small, we can simply discard this number (it doesn't really matter if it's there or not, since it's as small as we want it to be). Then, we get\[1 - 16x^3 > 0\]which finally yields\[x < \frac{1}{\sqrt[3]{16}}\]which is the point where the minimum of the function lies (it's the point where the function stops decreasing and starts increasing). Then, the minimum is\[f\left(\frac{1}{\sqrt[3]{16}}\right) = -\frac{3}{4\sqrt[3]{16^2}}\]so the range of the function is\[\left[-\frac{3}{4\sqrt[3]{16^2}}, \infty\right)\](this is far from a rigorous proof, but I really didn't find a better way without having to use "real" calculus).

Answer 14

I don't understand that value for f in the last line.

If I calculate it, I get about -0.4725 (yours is about -1/8).

http://www.wolframalpha.com/input/?i=plot+x^2-sqrt%28x%29

Answer 15

wow. i didn't mean to create so much work. some one asked this question here (it was in a work sheet) and i said i couldn't do it without calculus but it was a pre-calc course so i thought there was some precalc thing i was missing. i seriously doubt they expected all the work above. totally impressive though

Answer 16

\[y=x^2-\sqrt{x}\]
\[y-x^2=-\sqrt{x}\]
\[(y-x^2)^2=(-\sqrt{x})^2\]
\[y^2-2yx^2+x^4=x\]
\[y^2-2yx^2+x^4-x=0\]
\[x^4-2yx^2+y^2-x=0\]
Let u=x^2

\[u^2-2yu+(y^2-x)=0\]

\[u=\frac{2y \pm \sqrt{4y^2-4(y^2-x)}}{2}\]

we want 4y^2-4(y^2-x)>0
y^2-y^2+x>0
x>0
since the domain of f inverse is x>0
then the range of f is x>0

Answer 17

Um range is y values...

Answer 18

oops y>0

Answer 19

but still the graph is lower than 0
so this is not right

Answer 20

U had right before I think..

"satelitte, I have a question, is range from -0.47247 to infinity ?"

Answer 21

no but this was something similar to what i had seen a month or so ago. the "solve for x" thing which confused the bejesus out of me. i bet this is right

Answer 22

who knows? i didn't actually do it. i found the critical point but didn't bother plugging it in to get the decimal

Answer 23

because that was the calc way and i wanted a different way.

Answer 24

the critical point is
\[\frac{1}{2\sqrt[3]{2}}\] and i didn't feel like trying to evaluate the function there

Answer 25

and i still don't. but that is the answer, whatever it is
\[(\frac{1}{2\sqrt[3]{2}})^2-\sqrt{\frac{1}{2\sqrt[3]{2}}}\]

Answer 26

which is approximately -0.4724703927