`y = x^4 - 2x^2 + 3` Determine the point(s) (if any) at which the graph of the function has a horizontal tangent line.

Asked on by enotes

Textbook Question

Chapter 2, 2.2 - Problem 57 - Calculus of a Single Variable (10th Edition, Ron Larson).
See all solutions for this textbook.

1 Answer | Add Yours

zach9414's profile pic

zach9414 | High School Teacher | (Level 2) Adjunct Educator

Posted on

So I will start by rewriting this equation in the form:

`f(x)=x^4-2x^2+3 `

We know that if we have  then we can plug in any value for x, and the result will be what the slope of the function is at that given point. 

So if we find the derivative of this original function, we should be left with the following:

`f'(x)=4x^3-4x `  

If the tangent line to this graph is a horizontal line, then the slope should be 0/x or just 0.

So we can set what we calculated for f'(x) equal to zero and we can solve for our x-value:

`4x^3-4x=0 `

We should simplify this to get the following:

`4x(x^2-1)=0 `

`4x=0 `  and `(x^2-1)=0 `

`x=0 `  and `x-1=0 ` and `x+1=0 `

So x should equal -1, 0, and 1

So, this tells us that if our "x" value is zero, one or negative one, then the slope should be horizontal. Now that we know the x-values of the coordinates, lets plug these into the original equation to find the y-coordinates!:

`f(-1)=(-1)^4 -2(-1)^2+3=1-2+3=2 `

`f(0)=(0)^4-2(0)^2+3 = 0+0+3 = 3 `

`f(1)=(1)^4 - 2(1)^2 + 3 = 1-2+3 = 2`

` `

Therefore, our coordinates for the points where the tangent line to the graph are horizontal are:

(-1,2), (0,3), and (1,2)

We’ve answered 319,667 questions. We can answer yours, too.

Ask a question