Question
The curve y=2x^3 + ax^2 + bx has no turning points. Show that a^2 < 6b?
Answers
GPT-4o mini
To determine when the curve given by \( y = 2x^3 + ax^2 + bx \) has no turning points, we start by finding its derivative:
\[
y' = \frac{dy}{dx} = 6x^2 + 2ax + b
\]
Turning points occur where the derivative \( y' \) is zero, i.e., \( 6x^2 + 2ax + b = 0 \). For there to be no turning points, this quadratic equation in \( x \) must have no real roots. This happens when the discriminant of the quadratic is less than zero.
The discriminant \( D \) of the quadratic \( 6x^2 + 2ax + b \) can be calculated as follows:
\[
D = (2a)^2 - 4 \cdot 6 \cdot b = 4a^2 - 24b
\]
To ensure that there are no turning points, we require:
\[
D < 0
\]
This leads us to:
\[
4a^2 - 24b < 0
\]
We can rearrange this inequality:
\[
4a^2 < 24b
\]
Dividing both sides by 4 gives:
\[
a^2 < 6b
\]
Thus, we have shown that for the curve \( y = 2x^3 + ax^2 + bx \) to have no turning points, the condition \( a^2 < 6b \) must hold true.
\[
y' = \frac{dy}{dx} = 6x^2 + 2ax + b
\]
Turning points occur where the derivative \( y' \) is zero, i.e., \( 6x^2 + 2ax + b = 0 \). For there to be no turning points, this quadratic equation in \( x \) must have no real roots. This happens when the discriminant of the quadratic is less than zero.
The discriminant \( D \) of the quadratic \( 6x^2 + 2ax + b \) can be calculated as follows:
\[
D = (2a)^2 - 4 \cdot 6 \cdot b = 4a^2 - 24b
\]
To ensure that there are no turning points, we require:
\[
D < 0
\]
This leads us to:
\[
4a^2 - 24b < 0
\]
We can rearrange this inequality:
\[
4a^2 < 24b
\]
Dividing both sides by 4 gives:
\[
a^2 < 6b
\]
Thus, we have shown that for the curve \( y = 2x^3 + ax^2 + bx \) to have no turning points, the condition \( a^2 < 6b \) must hold true.