2. For what values of x dies the graph of f(x)=2x^3-3x^2-6x+87 have a horizontal tangent?
My answer:
(1.6,77.9)
(-0.62, 89.1)
dy/dx = 6x^2 - 6x - 6
= 0 for a horizontal slope
x^2 - x - 1 = 0
x = (1 ± √5)/2
= appr 1.618 or -.618
you are correct, but all it asked for is the values of x
(did you know that the first answer x = 1.618.. is the "golden ratio " ? )
Well, the graph of the function will have a horizontal tangent when its derivative is equal to zero. Let's find the derivative of f(x):
f'(x) = (6x^2 - 6x - 6)
Now we can solve for x:
6x^2 - 6x - 6 = 0
Divide both sides by 6:
x^2 - x - 1 = 0
Using the quadratic formula, we can find the solutions for x:
x = (-(-1) ± sqrt((-1)^2 - 4(1)(-1))) / (2(1))
x = (1 ± sqrt(5)) / 2
So the horizontal tangents occur at x = (1 + sqrt(5))/2 and x = (1 - sqrt(5))/2.
Calculating the corresponding y-coordinates using f(x), we get:
For x = (1 + sqrt(5))/2: f((1 + sqrt(5))/2) = 77.9
For x = (1 - sqrt(5))/2: f((1 - sqrt(5))/2) = 89.1
Therefore, the points where the graph of f(x) has a horizontal tangent are approximately (1.6, 77.9) and (-0.62, 89.1).
To find the values of x where the graph of f(x) = 2x^3 - 3x^2 - 6x + 87 has a horizontal tangent, we need to find the values that make the derivative of f(x) equal to zero.
First, let's find the derivative of f(x):
f'(x) = 6x^2 - 6x - 6
Next, set f'(x) equal to zero and solve for x:
6x^2 - 6x - 6 = 0
Divide both sides by 6 to simplify the equation:
x^2 - x - 1 = 0
There are two ways to solve this quadratic equation. One way is by factoring, and the other way is by using the quadratic formula. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -1, and c = -1. Substituting these values into the quadratic formula, we get:
x = (-(-1) ± √((-1)^2 - 4(1)(-1))) / (2(1))
Simplifying further:
x = (1 ± √(1 + 4))/2
x = (1 ± √5)/2
Therefore, the horizontal tangents occur at x = (1 + √5)/2 and x = (1 - √5)/2.
Approximating these values, we get:
x ≈ 1.62 and x ≈ -0.62
Now we need to find the corresponding y-values for these x-values by plugging them back into the original function:
f(1.62) ≈ 77.86
f(-0.62) ≈ 89.08
Therefore, the graph of f(x) = 2x^3 - 3x^2 - 6x + 87 has horizontal tangents at approximately (1.62, 77.86) and (-0.62, 89.08).
To find the values of x for which the graph of f(x) has a horizontal tangent, we need to find the points where the derivative of f(x) is equal to zero.
First, let's find the derivative of f(x):
f'(x) = d/dx(2x^3 - 3x^2 - 6x + 87)
= 6x^2 - 6x - 6
Now, set f'(x) equal to zero and solve for x:
6x^2 - 6x - 6 = 0
To solve this quadratic equation, we can use either factoring, completing the square, or the quadratic formula. In this case, let's use factoring:
6(x^2 - x - 1) = 0
Next, set each factor equal to zero:
x^2 - x - 1 = 0
Now, we can solve this quadratic equation. However, it cannot be factored easily, so let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = 1, b = -1, and c = -1. Substituting these values into the formula gives us:
x = (-(-1) ± √((-1)^2 - 4(1)(-1))) / (2(1))
x = (1 ± √(1 + 4)) / 2
x = (1 ± √5) / 2
Therefore, the solutions to the equation are x = (1 + √5) / 2 and x = (1 - √5) / 2.
Now we have the values of x where the derivative of f(x) is equal to zero. To find the points where the graph of f(x) has a horizontal tangent, we substitute these x-values back into the original function f(x):
For x = (1 + √5) / 2:
f((1 + √5) / 2) = 2((1 + √5) / 2)^3 - 3((1 + √5) / 2)^2 - 6((1 + √5) / 2) + 87
f((1 + √5) / 2) ≈ 77.9
For x = (1 - √5) / 2:
f((1 - √5) / 2) = 2((1 - √5) / 2)^3 - 3((1 - √5) / 2)^2 - 6((1 - √5) / 2) + 87
f((1 - √5) / 2) ≈ 89.1
Therefore, the graph of f(x) has a horizontal tangent at approximately (1.6, 77.9) and (-0.62, 89.1).