let f(x) = x^3 +1 a.find the points on the graph of f where the slope of the tangent line is equal 12
y' = 3x^2
so, if 3x^2 = 12, x = -2 or 2
The points are thus (-2,-7) and (2,9)
Why did the mathematician go to the party?
Because they heard there would be a lot of tangent lines in need of a slope!
Alright, let's find the points on the graph of f(x) = x^3 + 1 where the slope of the tangent line is equal to 12.
We can start by finding the derivative of f(x) to get the equation for the slope of the tangent line. The derivative of f(x) = x^3 + 1 is:
f'(x) = 3x^2
Now, we need to set the derivative f'(x) equal to 12:
3x^2 = 12
Dividing both sides of the equation by 3:
x^2 = 4
Taking the square root of both sides:
x = ±2
So, the points on the graph of f where the slope of the tangent line is equal to 12 are x = 2 and x = -2. We can substitute these values back into f(x) to find the corresponding y-coordinates:
For x = 2:
f(2) = (2)^3 + 1 = 8 + 1 = 9
So, the point is (2, 9).
For x = -2:
f(-2) = (-2)^3 + 1 = -8 + 1 = -7
So, the point is (-2, -7).
There you have it! The points on the graph of f(x) = x^3 + 1 where the slope of the tangent line is equal to 12 are (2, 9) and (-2, -7).
To find the points on the graph of f(x) = x^3 + 1 where the slope of the tangent line is equal to 12, we need to find the derivative of f(x) and set it equal to 12.
Step 1: Find the derivative of f(x)
To find the derivative of f(x), we'll use the power rule of differentiation. For a function f(x) = x^n, the derivative is given by f'(x) = nx^(n-1).
In this case, f(x) = x^3 + 1, so the derivative of f(x) is:
f'(x) = 3x^(3-1) = 3x^2.
Step 2: Set the derivative equal to 12
Since we want the slope of the tangent line to be equal to 12, we set the derivative equal to 12 and solve the equation:
3x^2 = 12.
Step 3: Solve the equation
To solve the equation 3x^2 = 12, divide both sides by 3:
x^2 = 4.
Take the square root of both sides:
x = ±2.
So, we have two possible x-values where the slope of the tangent line is 12: x = -2 and x = 2.
Step 4: Find the corresponding y-values.
To find the corresponding y-values, substitute the x-values into the original function f(x) = x^3 + 1.
For x = -2:
f(-2) = (-2)^3 + 1 = -8 + 1 = -7.
For x = 2:
f(2) = (2)^3 + 1 = 8 + 1 = 9.
The points on the graph of f where the slope of the tangent line is equal to 12 are (-2, -7) and (2, 9).
To find the points on the graph of f(x) where the slope of the tangent line is equal to 12, we need to find the points where the derivative of f(x) is equal to 12.
First, let's find the derivative of f(x). The derivative of f(x) can be found using the power rule, which states that the derivative of x^n is equal to n*x^(n-1). Applying the power rule to each term:
f'(x) = d/dx (x^3 + 1)
= 3x^(3-1)
= 3x^2
Now, we have the derivative of f(x), which is 3x^2. We want to find the points on the graph where the slope of the tangent line, which is given by the derivative, is equal to 12.
So, we set f'(x) = 12 and solve for x:
3x^2 = 12
Divide both sides by 3:
x^2 = 4
Taking the square root of both sides, we get:
x = ±√4
Simplifying, we find:
x = ±2
So, the x-values where the slope of the tangent line is equal to 12 are x = 2 and x = -2.
To find the corresponding y-values, we substitute these x-values back into the original function f(x):
For x = 2:
f(2) = (2^3) + 1
= 8 + 1
= 9
For x = -2:
f(-2) = (-2^3) + 1
= -8 + 1
= -7
Therefore, the points on the graph of f where the slope of the tangent line is equal to 12 are (2, 9) and (-2, -7).