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The function f(x) = x^3 - 3x + 2 is differentiable everywhere. Find its critical
The function f(x) = x^3 - 3x + 2 is differentiable everywhere. Find its critical points.
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The function f(x) = x^3 - 3x + 2 is differentiable everywhere. Find its critical points.
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f'(x) = 3x^2 - 3 = 0 gives x = ±1, thus critical points are x = -1 and x = 1.
Questions & Step-by-step Solutions
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Q: The function f(x) = x^3 - 3x + 2 is differentiable everywhere. Find its critical points.
Solution:
f'(x) = 3x^2 - 3 = 0 gives x = ±1, thus critical points are x = -1 and x = 1.
Steps: 8
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Step 1: Start with the function f(x) = x^3 - 3x + 2.
Step 2: Find the derivative of the function, which is f'(x). The derivative tells us the slope of the function.
Step 3: Calculate the derivative: f'(x) = 3x^2 - 3.
Step 4: Set the derivative equal to zero to find critical points: 3x^2 - 3 = 0.
Step 5: Solve the equation 3x^2 - 3 = 0. First, add 3 to both sides: 3x^2 = 3.
Step 6: Divide both sides by 3: x^2 = 1.
Step 7: Take the square root of both sides: x = ±1. This means x can be 1 or -1.
Step 8: The critical points are the values of x we found: x = -1 and x = 1.
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