Find the stationary point on the curve y=x^2-1 and show that it is a minimum point by checking the derivative on both sides of it.
I've found that its stationary point is (0, -1), just don't understand what "check the derivatives of both sides of it" means..
draw the curve. If it's a minimum at (0,-1) then
(a) y will have negative slope on the left: y' < 0
(b) y will have positive slope on the right: y' > 0
That is, y will be decreasing, hit the minimum, and then start increasing
FYI, if y" > 0 at x=0, then the graph is concave up, so it is a minimum there.
To find the stationary point on the curve y = x^2 - 1 and determine if it is a minimum point, we first need to understand the concept of derivatives and how they can help us.
The derivative of a function gives us information about the rate at which the function is changing at any given point. In the context of finding stationary points, we are interested in where the derivative is equal to zero. At these points, the function is not increasing or decreasing but rather remains "stationary."
To find the stationary point(s), we have to find the derivative of the given function, which, in this case, is y = x^2 - 1. The derivative is denoted as dy/dx and represents the rate of change of y with respect to x.
Taking the derivative of y = x^2 - 1 with respect to x using differentiation rules, we get:
dy/dx = 2x
Now, we can set dy/dx equal to zero to find the x-coordinate(s) of the stationary point(s):
2x = 0
Solving for x, we find:
x = 0
So, the x-coordinate of the stationary point is 0.
To determine if this stationary point is a minimum point, we can check the sign of the derivative on both sides of x = 0. Intuitively, if the derivative changes from negative to positive when approaching x = 0, it means the function is increasing and x = 0 corresponds to a minimum point. Similarly, if the derivative changes from positive to negative when approaching x = 0, it means the function is decreasing, and x = 0 corresponds to a maximum point.
Substituting values on both sides of x = 0 into the derivative equation, we have:
dy/dx = 2(0) = 0
Since the derivative is zero at x = 0, we need to examine the behavior of the function around this point. Plugging in values slightly greater and slightly smaller than x = 0 into the original function y = x^2 - 1, we have:
For x = -1 (slightly smaller than 0):
y = (-1)^2 - 1 = 1 - 1 = 0
For x = 1 (slightly greater than 0):
y = (1)^2 - 1 = 1 - 1 = 0
Observing the y-values, we see that the function goes from positive to negative as we move from x = -1 to x = 0 and then from negative to positive as we move from x = 0 to x = 1.
Since the derivative changes sign on both sides of x = 0, we can conclude that the function has a minimum point at x = 0. This means the stationary point is a minimum point on the curve y = x^2 - 1.
In conclusion, the stationary point on the curve y = x^2 - 1 is (0, -1), and we established it is a minimum point by checking the sign of the derivative on both sides of x = 0.