Which function has the most horizontal tangents?

A.f(x) = 3x^3
B.f(x)=x^4-1
C.f(x)=x^2-2x+1
D.f(x)=x^3+ 3x^2-9x-1

D has 2

The others all have 1

Don't forget your Algebra II now that you're taking calculus.

Well, I have to say, the function that has the most horizontal tangents is probably the one that needs the most support! So, I'd have to go with option D, f(x) = x^3 + 3x^2 - 9x - 1. It's definitely in need of some stability!

To determine which function has the most horizontal tangents, we need to find the derivative of each function and then count the number of times the derivative equals zero for each function.

Let's find the derivative of each function:

A. f(x) = 3x^3
Taking the derivative of f(x) with respect to x, we get:
f'(x) = 9x^2

B. f(x) = x^4 - 1
Taking the derivative of f(x) with respect to x, we get:
f'(x) = 4x^3

C. f(x) = x^2 - 2x + 1
Taking the derivative of f(x) with respect to x, we get:
f'(x) = 2x - 2

D. f(x) = x^3 + 3x^2 - 9x - 1
Taking the derivative of f(x) with respect to x, we get:
f'(x) = 3x^2 + 6x - 9

Next, we need to find the zeros of each derivative to determine the x-values where the tangents are horizontal:

A. Setting f'(x) = 0:
9x^2 = 0
x^2 = 0
This only has one solution: x = 0

B. Setting f'(x) = 0:
4x^3 = 0
x^3 = 0
This only has one solution: x = 0

C. Setting f'(x) = 0:
2x - 2 = 0
2x = 2
x = 1

D. Setting f'(x) = 0:
3x^2 + 6x - 9 = 0
Dividing by 3:
x^2 + 2x - 3 = 0
Factoring:
(x + 3)(x - 1) = 0
Setting each factor equal to zero:
x + 3 = 0 => x = -3
x - 1 = 0 => x = 1

Now, let's count the number of horizontal tangents for each function:

A. f(x) = 3x^3
This function has one horizontal tangent at x = 0.

B. f(x) = x^4 - 1
This function also has one horizontal tangent at x = 0.

C. f(x) = x^2 - 2x + 1
This function has one horizontal tangent at x = 1.

D. f(x) = x^3 + 3x^2 - 9x - 1
This function has two horizontal tangents at x = -3 and x = 1.

Therefore, the correct answer is D. function f(x) = x^3 + 3x^2 - 9x - 1 has the most horizontal tangents with two tangents.

To determine which function has the most horizontal tangents, we need to find the points on the graph where the derivative is equal to zero. The derivative represents the rate of change of a function at any given point.

Let's find the derivative for each function:

A. f(x) = 3x^3
To find the derivative, we can use the power rule. The power rule states that if we have a term of the form ax^n, the derivative will be n * ax^(n-1).
So, the derivative of f(x) = 3x^3 is: f'(x) = 3 * 3x^(3-1) = 9x^2.

B. f(x) = x^4 - 1
Using the power rule, the derivative of f(x) = x^4 - 1 is: f'(x) = 4x^3.

C. f(x) = x^2 - 2x + 1
The derivative of f(x) = x^2 - 2x + 1 can be found using the power rule:
f'(x) = 2x - 2.

D. f(x) = x^3 + 3x^2 - 9x - 1
Using the power rule, the derivative of f(x) = x^3 + 3x^2 - 9x - 1 is: f'(x) = 3x^2 + 2 * 3x - 9 = 3x^2 + 6x - 9.

To determine the points where the derivative is equal to zero, we need to solve each derivative equation for x.

A. Setting f'(x) = 9x^2 equal to zero and solving for x, we get:
9x^2 = 0
x^2 = 0
x = 0

B. Setting f'(x) = 4x^3 equal to zero:
4x^3 = 0
x^3 = 0
x = 0

C. Setting f'(x) = 2x - 2 equal to zero:
2x - 2 = 0
2x = 2
x = 1

D. Setting f'(x) = 3x^2 + 6x - 9 equal to zero:
3x^2 + 6x - 9 = 0
Using the quadratic formula or factoring, we can solve for x. However, for the purpose of this question, let's focus on the x-values.
The equation has two distinct solutions for x, which means there are two points on the graph where the derivative equals zero.

Now, let's count the number of horizontal tangents for each function:

A. f(x) = 3x^3: One horizontal tangent at x = 0.
B. f(x) = x^4 - 1: One horizontal tangent at x = 0.
C. f(x) = x^2 - 2x + 1: One horizontal tangent at x = 1.
D. f(x) = x^3 + 3x^2 - 9x - 1: Two horizontal tangents.

Therefore, the function with the most horizontal tangents is D. f(x) = x^3 + 3x^2 - 9x - 1, with two horizontal tangents.