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# We are going to be finding the absolute maxima and minima of a given function. # To do this we must take the derivative of our funtion, set it equal to zero, solve for x and then plug the value of x back in. We also must use the second derivative test to see if it is a maximum or minimum or use test values to find where it is increasing and decreasing. # 1. Given the function (below) find the derivative show(x^2 + 2*x + 4)
x2+2x+4\displaystyle x^{2} + 2 \, x + 4
derivative(x^2 + 2*x + 4)
2*x + 2
# 2. Set the derivative equal to 0 and solve for x. solve(2*x+2==0,x)
[x == -1]
# 3. Plug x=-1 back into the original function and find the x value or max/min at -1. x=-1 (x^2 + 2*x + 4)
3
# This means there is a maximum or a minimum at the point (-1,3). Now we have to find out whether it is a max or min by using the second derivative test or using test values, I am going to show you both ways. # 4. Using test values, take the x=-1 and pick one value below and one value above. We are going to use -2 and 0 as our test values. We then plug in the test values into the derivative to see if it is increasing or decreasing i.e. if the value is positive it is increasing and if the value is negative it is decreasing. # Lets test the lesser first and then the greater. x= -2 (2*x+2)
-2
# The value is negative meaning the function is decreasing as it approaches -1. # Now lets check the other side, using the greater value (0) x = 0 (2*x+2)
2
# This value is positive, meaning it is increasing as the function moves away from -1. # If the function decreases on one side of -1 and then increases on the other side of -1 then it is a minimum. # There is another way to find if it is a max or min, and that is using the second derivative test. That is, take the second derivative and if the value is less than 0 it is a max, greater than 0 it is a min. If it is equal to zero the test fails, and the test points should be used. # 5. # Here is our first derivative: x show(derivative(x^2 + 2*x + 4))
0
0\displaystyle 0
# Now, take the derivative again. diff(diff(x^2 + 2*x + 4,x))
0
# The second derivative of the function is positive 2. Going back to the rules stated above, this means that the point is a minimum because the second derivative is greater than zero. # Here is the function graphed with a dot at (-1,3). plot(x^2 + 2*x + 4,-10,10, ymin = 0, ymax = 10)+point((-1,3))