How many relative extrema can the polynomial t(x)=3x2−5x+8x3
have?(1 point)
The polynomial t(x) has two relative extrema.
To determine the number of relative extrema of a polynomial, we need to find the critical points. The critical points occur where the derivative of the polynomial equals zero or is undefined.
1. First, let's find the derivative of the given polynomial, t(x):
t'(x) = d/dx (3x^2 - 5x + 8x^3)
2. To find the derivative of each term, we can use the power rule:
t'(x) = 6x - 5 + 24x^2
3. Setting t'(x) equal to zero to find the critical points:
6x - 5 + 24x^2 = 0
4. Rearranging the equation:
24x^2 + 6x - 5 = 0
5. Solving the quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 24, b = 6, and c = -5.
x = (-6 ± √(6^2 - 4 * 24 * -5)) / (2 * 24)
x = (-6 ± √(36 + 480)) / 48
x = (-6 ± √516) / 48
6. Simplifying the square root:
x = (-6 ± 2√129) / 48
x = (-3 ± √129) / 24
Therefore, the critical points are (-3 + √129) / 24 and (-3 - √129) / 24.
Since we have found two distinct critical points, the polynomial t(x) can have at most 2 relative extrema.
To determine the number of relative extrema of a polynomial, we need to find its critical points. Critical points occur when the derivative of the polynomial equals zero or is undefined.
Let's start by finding the derivative of the polynomial t(x) = 3x^2 - 5x + 8x^3. Taking the derivative of each term separately, we get:
t'(x) = d(3x^2)/dx - d(5x)/dx + d(8x^3)/dx
Now, differentiate each term using the power rule:
t'(x) = 6x - 5 + 24x^2
To find the critical points, we need to solve the equation t'(x) = 0:
6x - 5 + 24x^2 = 0
Since this is a quadratic equation, we can solve it by factoring or using the quadratic formula. However, in this case, factoring is not possible.
Applying the quadratic formula, we have:
x = (-b ± √(b^2 - 4ac)) / 2a
For our equation, a = 24, b = 6, and c = -5.
x = (-6 ± √(6^2 - 4 * 24 * -5)) / 2 * 24
Simplifying further:
x = (-6 ± √(36 + 480)) / 48
x = (-6 ± √(516)) / 48
x = (-6 ± √(4 * 129)) / 48
x = (-6 ± 2√(129)) / 48
We have two solutions for x. Let's call them x1 and x2:
x1 = (-6 + 2√(129)) / 48
x2 = (-6 - 2√(129)) / 48
These are the values of x at which the derivative is equal to zero.
Next, we need to determine whether these critical points are relative extrema. To do this, we can use the second derivative test. Take the second derivative of the original polynomial t(x):
t''(x) = d/dx(6x - 5 + 24x^2)
t''(x) = 6 + 48x
Now, evaluate the second derivative at each critical point:
t''(x1) = 6 + 48 * [(-6 + 2√(129)) / 48] = 6 - 6 + 2√(129) = 2√(129) > 0
t''(x2) = 6 + 48 * [(-6 - 2√(129)) / 48] = 6 - 6 - 2√(129) = -2√(129) < 0
Since the second derivative is positive at x1 and negative at x2, we have one relative maximum at x1 and one relative minimum at x2.
Therefore, the polynomial t(x) = 3x^2 - 5x + 8x^3 has one relative extremum each at x1 and x2, for a total of 2 relative extrema.