Question

how many relative extrema can the polynomial t(x)=3x2−5x+8x3 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 derivative and then identify the critical points where this derivative equals zero.

1. **Find the derivative**:
\[
t'(x) = \frac{d}{dx}(3x^2 - 5x + 8x^3) = 6x - 5 + 24x^2
\]
This simplifies to:
\[
t'(x) = 24x^2 + 6x - 5
\]

2. **Set the derivative equal to zero** to find critical points:
\[
24x^2 + 6x - 5 = 0
\]

3. **Use the quadratic formula** to find the roots of the quadratic equation:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 24 \), \( b = 6 \), and \( c = -5 \):
\[
x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 24 \cdot (-5)}}{2 \cdot 24}
\]
\[
x = \frac{-6 \pm \sqrt{36 + 480}}{48}
\]
\[
x = \frac{-6 \pm \sqrt{516}}{48}
\]
\[
\sqrt{516} = \sqrt{4 \cdot 129} = 2\sqrt{129}
\]
Therefore:
\[
x = \frac{-6 \pm 2\sqrt{129}}{48} = \frac{-3 \pm \sqrt{129}}{24}
\]

4. **Calculate the number of real roots**:
The discriminant of the quadratic equation \( 6^2 - 4 \cdot 24 \cdot (-5) \) is positive (specifically, it's \( 516 \)), which means there are two distinct real roots for the equation \( 24x^2 + 6x - 5 = 0 \).

5. **Determine the nature of the critical points**:
Each critical point can correspond to either a relative maximum or relative minimum. Since the second derivative, \( t''(x) \), can help identify the nature of these critical points but is not crucial for counting relative extrema, we focus on the fact that there are 2 distinct real roots.

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