How do I find the relative maximum and minimum and the zeros of f(x)=-x^3+16x^2-76x+96??
since cubics are hard to solve, look for an easy root first:
-x^3+16x^2-76x+96
a quick synthetic division yields
-(x-2)(x^2-14x+48)
Now you have a quadratic which you can easily solve.
as for maxmima and minima, have you checked the first derivative yet? What did you find?
To find the relative maximum and minimum, and the zeros of the function f(x) = -x^3 + 16x^2 - 76x + 96, you can follow these steps:
Step 1: Find the derivative of the function.
The derivative of f(x) will help us analyze the critical points and determine where the function is increasing or decreasing.
f'(x) = -3x^2 + 32x - 76
Step 2: Set the derivative equal to zero and solve for x to find critical points.
To find the critical points, we need to solve the equation f'(x) = 0.
-3x^2 + 32x - 76 = 0
Using factoring or the quadratic formula, you can solve for x. In this case, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
Plugging in the values a = -3, b = 32, and c = -76:
x = (-32 ± √((32^2) - 4(-3)(-76)))/(2(-3))
x = (-32 ± √(1024 - 912))/(-6)
x = (-32 ± √112)/(-6)
x = (-32 ± 4√7)/(-6)
x = (16 ± 2√7)/3
The values of x are the critical points of f(x).
Step 3: Determine the nature of the critical points.
To determine whether the critical points are relative maximum, minimum, or neither, we will analyze the second derivative.
The second derivative is found by taking the derivative of the first derivative:
f''(x) = -6x + 32
Now we can evaluate f''(x) at each of the critical points.
For x = (16 + 2√7)/3:
f''((16 + 2√7)/3) = -6((16 + 2√7)/3) + 32
Plug this into a calculator to get an actual value, which is approximately 25.41.
Since f''((16 + 2√7)/3) is positive, the function is concave up at this critical point, indicating a relative minimum.
For x = (16 - 2√7)/3:
f''((16 - 2√7)/3) = -6((16 - 2√7)/3) + 32
Plug this into a calculator to get an actual value, which is approximately -1.74.
Since f''((16 - 2√7)/3) is negative, the function is concave down at this critical point, indicating a relative maximum.
Step 4: Find the zeros of the function.
To find the zeros of the function, we need to solve the equation f(x) = 0.
Setting f(x) = -x^3 + 16x^2 - 76x + 96 equal to zero and factoring, we get:
(x - 4)(x - 4)(x - 6) = 0
From this, we can see that there are two zeros at x = 4 and one zero at x = 6.
So, to summarize:
- The relative minimum occurs at x = (16 + 2√7)/3
- The relative maximum occurs at x = (16 - 2√7)/3
- The zeros of the function are x = 4 and x = 6.
I hope this helps! Let me know if you have any further questions.
To find the relative maximum and minimum and the zeros of the function f(x) = -x^3 + 16x^2 - 76x + 96, you need to follow a step-by-step process. Here's what you can do:
1. Find the derivative of the function: Differentiating f(x) will give you f'(x), which represents the slope of the function at any given point. In this case, f'(x) = -3x^2 + 32x - 76.
2. Set the derivative equal to zero and solve for x: For a relative extremum (maximum or minimum), the derivative is equal to zero. So, set f'(x) = 0 and solve for x. In this case, -3x^2 + 32x - 76 = 0.
3. Solve the equation from step 2 to find the critical points: Use factoring, completing the square, or the quadratic formula to solve for x. The solutions you obtain are the critical points of the function.
4. Test the critical points using the second derivative test: Take the second derivative, f''(x), to determine whether the critical points found in step 3 correspond to relative maxima or minima. If f''(x) > 0, it's a relative minimum; if f''(x) < 0, it's a relative maximum.
5. Find the zeros of f(x): Zeros of a function occur when f(x) = 0. To find the zeros, set f(x) = 0 and solve for x. The solutions you find will be the x-values at which the function crosses the x-axis.
By following these steps, you should be able to determine the relative maximum and minimum points as well as the zeros of the given function.