To find the maximum number of relative extrema a polynomial can have, we can use the derivative of the polynomial. The given polynomial is:
\[ t(x) = 8x^3 + 3x^2 - 5x \]
Step 1: Find the first derivative
The first derivative \( t'(x) \) is calculated as follows:
\[ t'(x) = \frac{d}{dx}(8x^3) + \frac{d}{dx}(3x^2) - \frac{d}{dx}(5x) \]
Calculating each term, we get:
\[ t'(x) = 24x^2 + 6x - 5 \]
Step 2: Set the derivative to zero
To find the relative extrema, we set the first derivative equal to zero:
\[ 24x^2 + 6x - 5 = 0 \]
Step 3: Use the quadratic formula
We can use the quadratic formula to solve for \( x \):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 24 \), \( b = 6 \), and \( c = -5 \).
Calculating the discriminant:
\[ b^2 - 4ac = 6^2 - 4(24)(-5) = 36 + 480 = 516 \]
Since the discriminant is positive (\( 516 > 0 \)), there are two distinct real roots.
Step 4: Determine relative extrema
Since \( t'(x) \) is a quadratic polynomial (degree 2), it can have at most 2 critical points (places where \( t'(x) = 0 \)), which could correspond to relative maxima or minima.
Conclusion
Thus, the polynomial \( t(x) = 8x^3 + 3x^2 - 5x \) can have at most 2 relative extrema.