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If h(x) = x^3 - 3x + 2, what is the critical point?
If h(x) = x^3 - 3x + 2, what is the critical point?
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Practice Questions
1 question
Q1
If h(x) = x^3 - 3x + 2, what is the critical point?
x = 0
x = 1
x = -1
x = 2
Show Solution
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h'(x) = 3x^2 - 3 = 0 gives x^2 = 1, so x = 1 and x = -1 are critical points.
Questions & Step-by-step Solutions
1 item
Q
Q: If h(x) = x^3 - 3x + 2, what is the critical point?
Solution:
h'(x) = 3x^2 - 3 = 0 gives x^2 = 1, so x = 1 and x = -1 are critical points.
Steps: 8
Show Steps
Step 1: Start with the function h(x) = x^3 - 3x + 2.
Step 2: Find the derivative of h(x), which is h'(x). The derivative tells us the slope of the function.
Step 3: Calculate the derivative: h'(x) = 3x^2 - 3.
Step 4: Set the derivative equal to zero to find critical points: 3x^2 - 3 = 0.
Step 5: Simplify the equation: 3x^2 = 3.
Step 6: Divide both sides by 3: x^2 = 1.
Step 7: Solve for x by taking the square root: x = 1 and x = -1.
Step 8: The critical points are x = 1 and x = -1.
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