use the second derivate test to locate the maxima and minima of y = x^2 + 2x - 3
since y'' = 2 any max/min will be a min.
what do you mean by that? :/
Take a look at the explanation of the 2nd derivative test. If y'' < 0 you have a max, and if y'' > 0 you have a min where y'=0.
if y' = 0 doesn't that mean there is no max or min? im sorry but i really don't understand this.. please help me
To locate the maxima and minima of a function using the second derivative test, we need to follow these steps:
Step 1: Find the first derivative of the function.
Step 2: Find the second derivative of the function.
Step 3: Set the second derivative equal to zero and solve for x to find any possible points of inflection.
Step 4: Determine the concavity of the function in the intervals between these points of inflection.
Step 5: Apply the second derivative test to determine the nature of any local maxima or minima.
Let's go through these steps for the given function, y = x^2 + 2x - 3:
Step 1: Find the first derivative.
Taking the derivative of y = x^2 + 2x - 3, we get y' = 2x + 2.
Step 2: Find the second derivative.
Taking the derivative of y' = 2x + 2, we get y'' = 2, which is a constant value.
Step 3: Set the second derivative equal to zero.
Since y'' = 2 is a constant, there is no critical point or point of inflection.
Step 4: Determine the concavity.
Since the second derivative is positive (2 > 0) and constant, the function is concave up for all values of x. This means there are no changes in concavity and no inflection points.
Step 5: Apply the second derivative test.
Since the second derivative is positive, there is no change in concavity, and no points of inflection, we conclude that the function has no local maxima or minima.
Therefore, the function y = x^2 + 2x - 3 has no local maxima or minima.