Question

How many relative extrema can the polynomial t(x)=3x^2-5x+8x^3 have?

Answer 1

To determine the number of relative extrema of the polynomial \( t(x) = 3x^2 - 5x + 8x^3 \), we first need to find its critical points by taking the derivative and setting it equal to zero.

1. **Find the derivative \( t'(x) \):**

\[
t(x) = 8x^3 + 3x^2 - 5x
\]
Using the power rule:
\[
t'(x) = \frac{d}{dx}(8x^3) + \frac{d}{dx}(3x^2) - \frac{d}{dx}(5x)
\]
\[
t'(x) = 24x^2 + 6x - 5
\]

2. **Set the derivative equal to zero to find critical points:**

\[
24x^2 + 6x - 5 = 0
\]

3. **Use the quadratic formula:**
The quadratic formula is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 24 \), \( b = 6 \), and \( c = -5 \).

Calculating the discriminant \( b^2 - 4ac \):
\[
b^2 - 4ac = 6^2 - 4 \cdot 24 \cdot (-5) = 36 + 480 = 516
\]

Now we can find the roots:

\[
x = \frac{-6 \pm \sqrt{516}}{2 \cdot 24} = \frac{-6 \pm \sqrt{516}}{48}
\]

Simplifying \( \sqrt{516} \):
\[
\sqrt{516} = \sqrt{4 \cdot 129} = 2\sqrt{129}
\]
Thus, the roots are:
\[
x = \frac{-6 \pm 2\sqrt{129}}{48} = \frac{-3 \pm \sqrt{129}}{24}
\]

4. **Determine the nature of the critical points:**
Since the derivative \( t'(x) \) is a quadratic function and it opens upwards (because the coefficient of \( x^2 \) is positive), it can have a maximum of 2 distinct real roots where the quadratic equals zero.

Given that the discriminant \( (b^2 - 4ac = 516) \) is positive, this means there are 2 distinct critical points.

5. **Conclusion:**
Each critical point can correspond to either a local minimum or a local maximum, but together, they represent two relative extrema.

Thus, the polynomial \( t(x) = 3x^2 - 5x + 8x^3 \) can have **two relative extrema**.