Suppose the function f is defined as follows: f(x)=4x^2-5x+1

1. Jun 29, 2015

Jaco Viljoen

1. The problem statement, all variables and given/known data
Suppose the function f is defined as follows:
f(x)=4x^2-5x+1
Write down the D:f and solve the inequality f(x)≤0
D:f (-∞,∞) because the function is a parabola, buy I am not sure how to solve the inequality, I would appreciate some help with this,

Thank you

2. Relevant equations
f(x)=4x^2-5x+1
f(x)≤0
3. The attempt at a solution
D:f (-∞,∞)
4x^2-5x+1≤0

Thank you

Last edited: Jun 29, 2015
2. Jun 29, 2015

PWiz

Why don't you substitute for f(x) in the inequality and see what you get? If you don't know how to deal with a quadratic inequality, try finding the critical values and then think about the graph of the function for a bit.

3. Jun 29, 2015

Jaco Viljoen

Had a dumb moment there, I am busy working on it now.

4. Jun 29, 2015

Jaco Viljoen

D:f (-∞,∞)
4x^2-5x+1≤0
(4x-1)(x-1)≤0
4x-1=0
4x=1
x=(1/4)

x-1=0
x=1

so x=1 and x=(1/4)
1/4≤x≤1

Thank you.

Jaco

Last edited: Jun 29, 2015
5. Jun 29, 2015

RUber

You have found the points where the function equals zero. Your intervals for the function being less than zero will either be between those points or outside of those points.
That is, your answer should be something like $1/4 \leq x \leq 1$ if between,
$x \leq 1/4 \text{ or } x \geq 1$ if outside.
One way to see what the proper interval is might be to test a few points.

Last edited by a moderator: Jun 29, 2015
6. Jun 29, 2015

Jaco Viljoen

Thank you Ruber,
I have mended my answer above,
Thank you again,

Jaco

7. Jun 29, 2015

RUber

I see you chose the correct interval, but did not express why it was correct. Did you plot a point inside or use a different method?

8. Jun 29, 2015

Jaco Viljoen

Ruben,
I substituted a few values into the equation see what was in fact true.
Thank you,

9. Jun 29, 2015

PWiz

A good way to solve these problems is to look at the coefficient of the x^2 term. If it is positive, you'll get a U shaped graph, and any inequality of the type "less than or equal to" will give you a closed interval, whereas an inequality of the type "greater than or equal to" will give you two separate inequalities. The reverse is true if the x^2 coefficient is negative (an inverted U shape graph is produced). This is of course only valid if a solution to the equality between the quadratic function and whatever it is being compared with exists.

10. Jun 29, 2015

Staff: Mentor

Strictly speaking, you were working with an inequality, not an equation.

11. Jun 29, 2015

HallsofIvy

Staff Emeritus
in addition to finding where $4x^2- 5x+ 1$ is equal to 0, you have determied that it can be factored as $(4x- 1)(x- 1)$ so the inequality can be written as $(4x- 1)(x- 1)\le 0$.

Now, use the fact that "the product of two numbers is negative if and only if the two numbers are of different sign".
Here, that tells you that either $4x- 1\le 0$ and $x- 1\ge 0$ or $4x- 1\ge 0$ and $x- 1\le 0$.