Graph this function.?

Question

can someone please graph this funtion and show me how it looks like?

f(x)= 12x^(2/3) - 4x

Answer 1

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Enter y = 12x^(2/3) - 4x

I tried it, closed up, it looks like a letter "n", with the start point of the "n" extending to positive infinity and the tail end of the letter "n" extending to negative infinity. There is an inflection point at the foot of the first leg of the letter "n".

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Answer 2

I presume you want to do the work, not just draw the graph.

First, f(0) = 0; f(1) = 8; f(-1) = 16. That gives us 3 easy points right away. Next, notice that x^(2/3) is the cube root of x^2, so x^(2/3) is never negative. Also, x^(2/3) will always be less than x, at least for sufficiently large positive x.

Another point we can get is f(8) = 12*8^(2/3) - 4*8 = 12*4 - 32 = 16. Then, f(-8) = 12*4 + 32 = 80.

For large positive x, the (-4x) term will dominate, so the function will go to negative infinity. For large negative x, the (-4x) term will also dominate, so the function goes to positive infinity.

Let's see if we can get the zeroes. (We already have f(0)=0.)
Set f(x) = 12x^(2/3) - 4x = 0. Dividing out the 4, that factors into x = 0 and

3x^(-1/3) - 1 = 0
3/x^(1/3) = 1
x^(1/3) = 3

Cube both sides to get x = 27.

That ought to be a root. Let's check:

f(27) = 12(27)^(2/3) -4(27)
f = 12(3^2) - 108 = 108 - 108 = 0

The roots of the function are at (0,0) and (27,0).

Now let's look for relative max/min points. Take the derivative:

df/dx = 12(2/3)x^(-1/3) - 4 = 0
8x^(-1/3) = 4
x^(1/3) = 8/4 = 2
x = 8

should be a maximum. But we already have f(8) = 16.

At this point I graph the function using these points:

(-8,80), (-1,16), (0,0), (1,8), (8,16)(maximum), (27,0)

The only thing that bothers me is that we didn't get a minimum at (0,0). Maybe the derivative (slope) is discontinuous there.

And in fact, it is. For the derivative, we had

df/dx = 12(2/3)x^(-1/3) - 4

which has a divide by zero at x=0.

Last thing. We ought to check the behavior of the function in the domain -1
f(1/8) = 12(1/8)^(2/3) - 4(1/8) = 12/4 - 1/2 = 5/2

f(-1/8) = 12(-1/8)^(2/3) - 4(-1/8) = 3 + 1/2 = 7/2

And there's two more points to plot.

We now have enough to plot the function confidently. It comes in from negative infinity with a mild slope through the third quadrant, curves down to the origin sharply; then starts up gently in the first quadrant to a flattened-out maximum at (8,16); then gently goes down the hill to a root at (27,0), gradually continuing on down toward negative infinity through the fourth quadrant.

There's your answer.

Answer 3

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