Question

Hi! What are the coordinates of the points at which the slope of the function with the equation y = x^4 - 8x^2 = 0?

Answer 1

i.e. take the derivative, set it equal to zero and solve right?

Answer 2

\[y=x^3-8x^2\]
\[y'=4x^3-16x=4x(x^2-4)\] solve
\[x(x^2-4)=0\] for \(x\)

Answer 3

If we graph this we can more accurately see what is going on: So this from Wolfram Alpha: http://www.wolframalpha.com/input/?i=+y+%3D+x%5E4+-+8x%5E2

Answer 4

In this case, @satellite73 is right about taking the derivative and solving for when it equals 0.

If you look at the graph, and given insight about derivatives, I believe that you can see that you will wanting the coordinates of the lowest points of the two "humps" you see on either side of the origin.

Answer 5

*will be wanting

Answer 6

Here you are using the power rule to get the derivative, which can tell you the "slope" - the change in y with regard to x.

The power rule is useful for the type of equation you are working with: here is the general concepts:

\[ f(x) = x^n, f'(x) = n * x^{(n-1)} \]

Answer 7

Here is a graph of the derivative:

http://www.wolframalpha.com/input/?i=x%28x2%E2%88%924%29

Answer 8

You see that this function intersects the x-axis 3 times, meaning there are 3 values for x where y = 0. So you will need to have 3 answers to your problem

Answer 9

hi! I am getting (0,0), (8^.5, 0), (-8^.5, 0). I think this is right but my stupid worksheet has no way to answer with an exponent/sq. root. but thank you all!

Answer 10

\[x(x^2-4)=0\] Gives you x=0, then \[x^2=4\] so x = {-2, 0, 2}

Answer 11

ah OK thanks flan ! i think I got it.

Answer 12

Then if you plug in those values of x, you obtain the points from the original equation

Answer 13

Nice, yeah, you should have it now ;)