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.
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Find the derivative \( t'(x) \):
\[ t(x) = 8x^3 + 3x^2 - 5x \] Using the power rule: \[ t'(x) = \frac{d}{dx}(8x^3) + \frac{d}{dx}(3x^2) - \frac{d}{dx}(5x) \] \[ t'(x) = 24x^2 + 6x - 5 \]
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Set the derivative equal to zero to find critical points:
\[ 24x^2 + 6x - 5 = 0 \]
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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: \[ x = \frac{-6 \pm 2\sqrt{129}}{48} = \frac{-3 \pm \sqrt{129}}{24} \]
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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.
- 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.