# Thread: Confused by One-to-one problems...

1. ## Confused by One-to-one problems...

The problem is...

Let f(x)= 5x^3 + kx^2 + 5x, k a constant. For what values of k is f one-to-one?

How would you work this problem to find what k is?

Thank You

2. Originally Posted by czaba
The problem is...

Let f(x)= 5x^3 + kx^2 + 5x, k a constant. For what values of k is f one-to-one?

How would you work this problem to find what k is?

Thank You
It means that $f(x)$ stricly increasing or decreasing.

$f'(x) = 15x^2 +2kx+5$.

To be strictly increasing or decreasing we require that this derivative be either always positive or negative. Since the leading coefficient is positive it means this must be strictly increasing. This quadradic is positive when the discrimant is negative,
$(2k)^2 - 4(15)(5) <0$
Solve for $k$.

3. How did you get that from 5x^3 + kx^2 + 5x ?

4. Originally Posted by czaba
How did you get that from 5x^3 + kx^2 + 5x ?
$(5x^3+kx^2+5x)' = 5(x^3)'+k(x^2)'+5(x)' = 15x^2+2kx+5$.

5. Okay... but why did we take the derivative?

6. Originally Posted by czaba
Okay... but why did we take the derivative?
Because if a differenciable function is one-to-one it means (meaning it satisfies the horizontal line test) then it must be either increasing or decreasing. Because if it makes a turn then the horizontal line test will hit two point there and so it is no longer one-to-one. Which means it must be either increasing or decreasing. Now the sign of the derivative tells if a function is increasing or decreasing. In this case the derivative was a quadradic function. Since the coefficient of this quadradic function is positive it means the function cannot be always negative, i.e. it cannot be decreasing. Thus, the only possible situation is when that quadradic function has discrimant < zero (that is where you get complex zeros).

7. Im sorry but I still dont understand how we got from the derivative to $
(2k)^2 - 4(15)(5) <0
$

8. Originally Posted by czaba
Im sorry but I still dont understand how we got from the derivative to $
(2k)^2 - 4(15)(5) <0
$
Let $ax^2+bx+c$ be a quadradic function. Now the rule is if the discrimanant which is the expression $b^2 - 4ac$ is negative then this parabola never intersects the x-axis. Which means it is either always positive or always negative. Meaning it has always one sign. This is what we want. Because we want the function to be increasing or decreasing. Which means the derivative is either always positive or negative. And this precisely happens when the discrimant is negative.