Calculus Help?!?

Question

Use the First derivative test to find the critical values of y=(x^(2)-4)^7. Use the First Derivative Test to determine if there are extrema and, if so, what kind?

Answer 1

Well, since the problem asks you to use the first derivative test, that's usually a good indicator that you should start by finding the first derivative. We have:

dy/dx = d((x² - 4)^7)/dx
= 7(x² - 4)^6 d(x²-4)/dx
= 14x(x² - 4)^6

Now, we need to find the critical points. These are the points where the derivative is either zero or nonexistent. Since the derivative exists everywhere, we only need to find the points where it is zero, so setting the derivative equal to zero and solving:

14x(x² - 4)^6 = 0
x=0 ∨ x² - 4 = 0
x = 0 ∨ x = 2 ∨ x = -2

Now, to find the type of critical point, we need to know whether the function is increasing or decreasing near the point. At x=0, note that 14(x²-4)^6 is positive in a neighborhood of zero, so 14x(x²-4)^6 is positive just to the right of zero and negative just to the left of zero. Thus by the first derivative test, we have that y is decreasing just to the left of the point and y is increasing just to the right of the point, so zero is a local minimum. At x=2, we have that 14x is positive in a neighborhood of 2, and (x²-4)^6 is positive on both sides of 2, so y' is positive near 2 on both sides, so by the first derivative test, 2 is an inflection point of the function. Conversely, 14x(x²-4)^6 is negative near -2 on both sides, so -2 is also an inflection point.

Answer 2

All i get is the dy/dx = 7((x^2)-4)^6 * (2x)

Type it into the calculator and look at the graph to come up with the extremas

Answer 3

Here's a little something to get you started:

y=(x^(2)-4)^7

y'=(x^2-4)^6 * 2x * 7

Answer 4

You have to get a calculator