# The Function f(x) = 6x^2+5x

1. Oct 20, 2015

### Jaco Viljoen

Good afternoon everyone,
I am preparing for exams and thought I would post a couple old paper questions and answers for what I have done,
As always I appreciate any input and feedback.

1. The problem statement, all variables and given/known data
The Function ƒ(x) = 6x^2+5x
a) Write down the Dƒ
b) Solve the inequality f(x) < 4

2. Relevant equations

3. The attempt at a solution
a)x∈ℝ
b)

6x^2+5x<4
6x^2+5x-4<0
(3x+4)(2x-1)<0

3x+4<0
3x<-4
x<-(4/3)

2x-1<0
2x<1
x<1/2

(-4/3)< x <(1/2) I am not sure why x is greater than -4/3 why would the sign change direction here?

Last edited: Oct 20, 2015
2. Oct 20, 2015

### RUber

You found that the (only) zeros of the function are at -4/3 and 1/2. So between those points the function will have a constant sign (negative or positive) and outside of those points the sign will be opposite what it is between them.
All you need to know is the sign of one point, (not one of the zeros), and you know the signs in all those intervals.
x=0 is between -4/3 and 1/2.
At x = 0, f(x) =0, so f(x)<4 is true.
Therefore the only interval that solves the inequality is between the zeros of f(x)-4 = 0, which you already know.

In your work, separating the inequalities, you missed the important feature of inequalities of products.
if (a)(b)<0, then either a is negative and b is positive or b is negative and a is positive, but both negative or both positive would give a positive product.

3. Oct 20, 2015

### Jaco Viljoen

So where in my work did I go wrong?
I am not sure I see the problem, would you mind showing me?

Thank you RUber

4. Oct 20, 2015

### Staff: Mentor

Above, you are missing ' < 0' in the last inequality.
Both of the above are wrong for the reason that RUber gives. In order for (3x + 4)(2x - 1) < 0, exactly one of the following must be true:
1. 3x + 4 < 0 AND 2x - 1 > 0,
OR
2. 3x + 4 > 0 AND 2x - 1 < 0

Investigate each of the two cases above. (You'll find that one of them can't happen.)

5. Oct 20, 2015

### Ray Vickson

Think of the graph y = f(x). You want to know the x-region where y < 4, which is the part below the horizontal line at height 4. The graph crosses the line at two points, which you have already found. Between these two roots the function f(x)-4 has one sign, and beyond the roots (in either direction) it has the opposite sign. To figure out which is which, just look at any particular value, such as f(0) - 4, to see which of the three regions the point x = 0 lies within. From that you can easily work out the complete solution.

Alternatively, you can take a more "algebraic" approach and examine the factors of the function f(x)-4, which you have done, but incorrectly.