Sketch the graph y=x^2 and y=x^4. what is the max vertical distance between the graphs??

Question

I have asked this question, yet i still don't understand?? please someone can you thoroughly explain it.

Answer 1

Get g(x) = x^4 - x^2.
Then you need to get the maximum of g(x). If you can use a graphics calculator to see the graph, there is no maximum that can be observed. g(x) is going up indefinitely. x^4 increases very much faster than x^2.

You can get the minimum distance, they both pass through (0,0) thus the min. distance is 0. No maximum.


d:

Answer 2

Define g(x) = x^4 - x^2, the vertical distance from the graphs y=x^2 and y=x^4, at a given x.

Then you want to find the maximum value of g.

g'(x) = 4x^3 - 2x = 0

=>

2x(2x^2 - 1) = 0

=>

x = 0 or x = +/- sqrt(2)/2

g(0) = 0 (so this gives the minimum vertical distance)

g(sqrt(2)/2) = 1/4 - 1/2 = -1/4 (we count this as 1/4, since we care about distance, which is positive)

g(-sqrt(2)/2) = 1/4 - 1/2 = -1/4 (same thing here)

So the max vertical distance is 1/4, and it occurs when x = +/- sqrt(2)/2.

**EDIT**

The guy below is right. The function g(x) appearing in my solution is only going to be valid for -1 <= x <= 1, otherwise there is no maximum vertical distance.

Answer 3

When x = 1, x^2 = 1 and x^4 = 1. The difference is 1-1 = 0.
When x = 2, x^2 = 4 and x^4 = 16. The difference is 16-4=12
As x gets bigger and bigger, the difference will approach infinity.

The problem makes more sense if you're restricted to x-values between -1 and 1. In that case, you're trying to find the maximum value of x^2 - x^4.

The maximum will be where the derivative has a slope of zero. That is, where 2x - 4x^3 = 0
Dividing by x: 2 - 4x^2 = 0
Rearranging: 4x^2 = 2
x^2 = 2/4 = 1/2
x = sqrt(0.5) or -sqrt(0.5)

To find the vertical distance, plug into your equation:
(sqrt(0.5))^2 - (sqrt(0.5))^4
0.5 - (0.5)^2
0.5 - 0.25
0.25

If you restrict x to be between -1 and 1, the maximum vertical distance between the graphs will be 0.25.

Hope that makes sense to you!