Finding a counterexample

[Username101]

Hi there,
Looking for some guidance in answering this question.

• Find a counterexample to show that the statement

2 x² + 3x + 1 ≥ 0

• I used
  

Let x = -1
and I still get 0

• I also tried -(1/3)

And I still get a positive fraction.



Any help would be appreciated.
Thanks!

 

[Deleted member 4993]

Hi there,
Looking for some guidance in answering this question.

• Find a counterexample to show that the statement

2 x² + 3x + 1 ≥ 0

• I used
  

Let x = -1
and I still get 0

• I also tried -(1/3)

And I still get a positive fraction.



Any help would be appreciated.
Thanks!

Plot

y = 2x² + 3x + 1

Observe where the x-intercepts are and observe that the values of y in that domain is < 0.

Choose one of those values (in between x intercepts).

 

[Username101]

Plot

y = 2x² + 3x + 1

Observe where the x-intercepts are and observe that the values of y in that domain is < 0.

Choose one of those values (in between x intercepts).

Hi thanks for your response!
So I'm solving the quadratic to find the x values then?

 

[Ishuda]

Hi there,
Looking for some guidance in answering this question.

• Find a counterexample to show that the statement

2 x² + 3x + 1 ≥ 0

• I used
  

Let x = -1
and I still get 0

• I also tried -(1/3)

And I still get a positive fraction.



Any help would be appreciated.
Thanks!

Another way: You have a place where it is zero so you know on of the factors? What is the other, i.e. divide (2 x² + 3x + 1) by (x+1) to get (2x+1) so you know
2 x² + 3x + 1 = (x+1) (2x+1)
Now 2x+1 is negative right around x=-1 but what is x+1.

 

[Deleted member 4993]

Hi thanks for your response!
So I'm solving the quadratic to find the x values then?

Did you plot the function?

 

[Username101]

Hi thanks for the replies.
No I didn't plot the function instead I used the quadratic formula

My results were -1 and -1/2

 

[stapel]

You've posted this to "Calculus", so you've already taken algebra. This exercise requires only algebra.

Find a counterexample to show that the statement

I will assume that you are supposed to find a counter-example to show that the statement is false.

. . . . .\(\displaystyle 2x^2\, +\, 3x\, +\, 1\, \geq\, 0\)

As you learned back in algebra, a positive quadratic graphs as an upward-opening parabola. What did you see when you did the graph?

This type of curve has its minimum at its vertex. You can complete the square (or else use the derivative) to find the x-coordinate of that vertex. Plugging this x-value into the quadratic will provide you with the minimum value of the quadratic. What did you get when you found the vertex?

If you use the derivative, you can find the x-value for the minimum. What did you get when you found the critical point? What did you get when you plugged this x-value into the quadratic?

Please be complete. Thank you!