find point(s) on the graph of the function f(x) = x^3 - 2 where the slope is 3

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Why did the function go to therapy? Because it couldn't find its slope's identity! Let's solve this mathematical mystery for f(x) = x^3 - 2. We need to find the point(s) where the slope is 3.

To find the slope, we need to take the derivative of f(x), which is f'(x) = 3x^2. Now, let's set f'(x) equal to 3 and solve for x:

3x^2 = 3

Divide both sides by 3:

x^2 = 1

Taking the square root of both sides, we get:

x = ±1

So, there are two points on the graph of f(x) = x^3 - 2 where the slope is 3: (1, -1) and (-1, -1). Be careful not to slip on these points, they can be quite slippery!

To find the point(s) on the graph of the function f(x) = x^3 - 2 where the slope is 3, we first need to find the derivative of the function and then solve for the x-values where the slope is equal to 3.

Step 1: Find the derivative of f(x)
The derivative of f(x) = x^3 - 2 can be found using the power rule. Taking the derivative of each term separately, we get:

f'(x) = 3x^2

Step 2: Set the derivative equal to 3
Since we want to find the x-values where the slope is equal to 3, we set the derivative f'(x) equal to 3:

3x^2 = 3

Step 3: Solve for x
To solve the equation 3x^2 = 3, we divide both sides by 3:

x^2 = 1

Taking the square root of both sides gives us:

x = ± 1

Step 4: Find the corresponding y-values
Now that we have the x-values where the slope is 3 (x = ±1), we can find the corresponding y-values by substituting these x-values back into the original function f(x) = x^3 - 2:

When x = 1:
f(1) = 1^3 - 2 = -1

When x = -1:
f(-1) = (-1)^3 - 2 = -3

Therefore, the points on the graph of the function f(x) = x^3 - 2 where the slope is 3 are (1, -1) and (-1, -3).

f' = 3x^2 - 2

3x^2 - 2 = 3
x^2 = 5/3

x = +/- sqrt (5/3)