use the derivative of the function
y=f(x)to find the points at which f has a
local maximum
local minimum
point of inflection
y'=(x-1)^2(x-2)(x-4)
if f(x)=uvw
then
f'(x)=uvw'+ uwv'+ wvu'
I will assume you really meant to write the derivative y' and not the function y although I would have thought the book would ask you to find that derivative.
Wherever that derivative, y', is zero, the curve is horizontal
This is twice at x = 1 and at x =2 and at x = 4
Look at the second derivative to see what happens there.
y'=(x-1)^2 (x^2-6x+8)
so
y" = (x-1)^2(2x-6)+2(x-1)(x^2-6x+8)
now what is y"at x = 1?
y"(1) = Zero, so that is an inflection point
what is y" at x = 2?
y"(2) = -2+0 = -2
That is a maximum because y" is negative
what is y"(4)?
y"(4) = 9(2)+2*3(16-24+8)
y"(4) = 18+0 = 18
so it is Minimum at x = 4
y"
To find the points at which the function has a local maximum, local minimum, or point of inflection, we need to make use of the first and second derivative tests.
1. Local Maximum:
To find the local maximum points of a function, we need to locate the critical points where the derivative equals zero or is undefined. So, we first find the derivative of the function:
y' = (x-1)^2(x-2)(x-4)
Next, we equate the derivative to zero and solve for x to find the critical points:
0 = (x-1)^2(x-2)(x-4)
Solving this equation, we find that the critical points occur at x = 1, x = 2, and x = 4.
Now, we analyze the nature of these critical points by examining the sign of the derivative on either side of each point. For instance, when x < 1, the derivative is negative, and when x > 1, the derivative is positive. Therefore, we can conclude that at x = 1, we have a local maximum point.
2. Local Minimum:
Similarly, we can find the local minimum points by identifying the critical points where the derivative equals zero or is undefined. Using the same derivative, we set it equal to zero:
0 = (x-1)^2(x-2)(x-4)
Solving this equation, we find the critical points at x = 1, x = 2, and x = 4. By analyzing the derivative's sign on either side of each point, we observe that when x < 2, the derivative is positive, and when x > 2, the derivative is negative. Therefore, we can conclude that at x = 2, we have a local minimum point.
3. Points of Inflection:
To find the points of inflection, we need to examine the concavity of the function. For this, we need to find the second derivative.
Taking the derivative of the previously found derivative y', we obtain the second derivative:
y'' = 2(x-1)(x-2)(x-4) + (x-1)^2(x-4) + (x-1)^2(x-2)
To find the points of inflection, we set the second derivative equal to zero:
0 = 2(x-1)(x-2)(x-4) + (x-1)^2(x-4) + (x-1)^2(x-2)
Simplifying and solving this equation will provide the x-values representing the points of inflection.
By applying these steps to the given function and analyzing each point, we can determine the local maximum, local minimum, and points of inflection.