Find the critical points of the linear equation y = 4x^3 - 3x

Question

So to be clear, when I state critical point, I want the maxima, minima, and any zeros in the domain of the function.

Answer 1

I wanted to make sure this was posted because in my session with one of my students, we had to stop mid-explanation due to a professional reason, so lets get to this:

Quadratic linear equations at some point cross the x and y axis, and quite simply its good to know a few key points: all x-intercepts including origin, the points it will reach before the domain (x values) result in a turn down/up (these are called the maxima and the minima).

To find the zeros first, we will use the original equation (y) and follow the steps below:

Set the equation equal to zero

factor out any x's from the terms

make conclusions about the terms remaining by setting each set of terms equal to zero

Step 1 is done for us.

y=4x^3 - 3x

Step 2:

y=4x^3 - 3x - we have an x in each term on the right side of the equation

y = x(4x^2 - 3) - now we have two terms that include an operation with x. (x) and (4x2-3)

Step 3:

Operationally this lets find what would make y in this linear equation equal 0.

It could be 0 => y=(0)(4x2 - 3) would make the final product zero

It could be making the difference between (4x2 - 3) equal zero.

so we do the following:

4x2 -3 = 0

4x2 = 3

x2 = 3/4

x = ±√(3)/2

+√(3)/2 and -√(3)/2 can both result in that part of the equation equal zero.

Then we apply what we know about these equations....the amount of maxima and minima in quadratic equations is a ((highest power exponent)-1). In this equation, it would be 2.

So x could be +√(3)/2, to make y = zero, and are CRITICAL POINTS in the domain.

These are the Zeros in the equation.

And the minimum x could equal because it is a quadratic equation with x raised to the 3rd power is -∞, and the maximum it could be is +∞.

Final answer pt 1 of 2:

The critical points in this problem are ,(0,-√(3)/2),(0,0), (0,+√(3)/2, +∞) per defining the domain. If the question needs the maxima or minima of the equation we can continue using original equation,y, and the first derivative of the original equation, y'.

Other textbooks and such define the critical points by including the maxima and minima points that a quadratic equation can be, and this would be found by taking the derivative of the linear equation, and then setting this new equation equal to zero.

So if y = 4x3 - 3x...applying the rules of the 1st derivative...

y' = 12x2 - 3

Then we set y' equal to zero.

12x2-3 =0

12x2=3

x2 = 3/12 = 1/4

x = ±√(1/4)

x = -(1/2) and +(1/2)

Then we plug these values back into the y, NOT y', to find the x coordinate

y = 4(1/2)3- 3(1/2)

y = 4(1/8) - 3/2

y= 4/8 - 3/2

y = 1/2 - 3/2

y= 2/2 = 1 so one point, the maxima, is (1/2,1).

y= 4(-1/2)3 - 3(-1/2)

y= 4(-1/8) + 3/2

y = (-4/8) + 3/2

y = -1/2 + 3/2

y = 2/2 = 1 so the other point, the minima, is (-1/2,1).

When you see a graph of such a scenario it makes sense as to what points are the focus.

Final answer pt. 2 ( if asked for) is :

The critical points in this problem are ,(√(3)/2,0),(0,0), (+√(3)/2,0)per defining the domain.

Maxima = (1/2,-1) and Minima (-1/2,1) when plugging in the zeros from the first derivative (y') into the original equation (y).

Answer 2

I hope this helps! Let me know if you have questions!

Answer 3

Uusually, the y-intercept is also considered a critical point. In this case, it is the origin, and it is also an x-intercept.

One major use of locating these points is to sketch the function. The end behaviors are also helpful for this purpose, as are locating any asymptotes.

Answer 4

y' = 12x² - 3 = 0; x = ±1/2. f(1/2) = -1; f(-1/2) = 1.

So, (1/2, -1) and (-1/2, 1) two critical points of this function