How many relative extrema can the polynomial t(x)=3x2−5x+8x3
have?(1 point)
at most
At most, the polynomial can have 2 relative extrema.
To determine the number of relative extrema of a polynomial, we need to find its derivative, then determine the number of solutions for which the derivative is equal to zero.
Finding the derivative of the polynomial t(x)=3x^2−5x+8x^3:
t'(x) = 6x - 5 + 24x^2.
To find the solutions for which t'(x) = 0, we can set the derivative equal to zero and solve for x:
6x - 5 + 24x^2 = 0.
Now, we can solve the quadratic equation by factoring or using the quadratic formula:
24x^2 + 6x - 5 = 0.
Using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a.
a = 24, b = 6, c = -5.
Substituting the values into the formula:
x = (-6 ± √(6^2 - 4*24*-5)) / (2*24).
Simplifying the equation:
x = (-6 ± √(36 + 480)) / 48,
x = (-6 ± √516) / 48,
x = (-6 ± 22.72) / 48.
This gives us two possible solutions for x:
x₁ = (-6 + 22.72) / 48 = 0.32,
x₂ = (-6 - 22.72) / 48 = -0.59.
Since we have two unique solutions, this means the polynomial t(x) has at most two relative extrema.
To find the number of relative extrema for a polynomial, we need to calculate the derivative of the polynomial and examine its critical points.
1. First, calculate the derivative of the polynomial: t’(x) = 6x - 5 + 24x^2
2. Set the derivative equal to zero to find critical points:
6x - 5 + 24x^2 = 0
3. Solve the above equation for x using factoring, the quadratic formula, or any other applicable method. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 24, b = 6, and c = -5. Plugging in these values, we get:
x = (-6 ± √(6^2 - 4 * 24 * -5)) / (2 * 24)
x = (-6 ± √(36 + 480)) / 48
x = (-6 ± √516) / 48
4. Simplify the expression:
x = (-6 ± √516) / 48
x = (-6 ± 2√129) / 48
x = (-3 ± √129) / 24
We have found two critical points: (-3 + √129) / 24 and (-3 - √129) / 24
5. Now, we need to determine whether these critical points are relative extrema or not. To do this, we can examine the second derivative of the polynomial.
The second derivative of the polynomial is: t''(x) = 6 + 48x
6. Plug the critical points into the second derivative:
t''((-3 + √129) / 24) = 6 + (48 * (-3 + √129) / 24)
t''((-3 - √129) / 24) = 6 + (48 * (-3 - √129) / 24)
If the second derivative at a critical point is positive, it corresponds to a relative minimum. If it is negative, it corresponds to a relative maximum. If it is zero, the test is inconclusive.
By evaluating the second derivative at both critical points, we can determine the number of relative extrema.
Note: It appears that there is an incomplete sentence in your original question ("at most"). If you meant to ask for the maximum number of relative extrema, we will proceed with the calculations. If you meant something else, please clarify your question.