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What are the critical points of the function f(x) = 2x^3 - 3x^2 + 4x - 1, and what type of points are they?
What are the critical points of the function f(x) = 2x^3 - 3x^2 + 4x - 1, and what type of points are they?
The critical points of the function are where the derivative is equal to zero or undefined. Taking the derivative of f(x), we get f'(x) = 6x^2 - 6x + 4. Setting this equal to zero, we get x = 1/2, which is a critical point. Since f''(1/2) = 12 > 0, this critical point corresponds to a relative minimum.
Using the Fundamental Theorem of Calculus, what is the value of the integral ∫(2x + 1) dx from x = 0 to x = 3?
Using the Fundamental Theorem of Calculus, what is the value of the integral ∫(2x + 1) dx from x = 0 to x = 3?
The Fundamental Theorem of Calculus states that if F(x) is an antiderivative of f(x), then ∫f(x) dx from a to b = F(b) - F(a). In this case, an antiderivative of 2x + 1 is x^2 + x, so ∫(2x + 1) dx from x = 0 to x = 3 = (3^2 + 3) - (0^2 + 0) = 12.
What is the displacement of a car that has a velocity of v(t) = t^2 - 6t + 8 during the time interval [1, 5] seconds?
What is the displacement of a car that has a velocity of v(t) = t^2 - 6t + 8 during the time interval [1, 5] seconds?
The displacement of the car during the time interval [1, 5] seconds is equal to the area under the velocity curve between t = 1 and t = 5. Integrating v(t) from 1 to 5, we get ∫(t^2 - 6t + 8) dt from 1 to 5 = [(5^3)/3 - 15(5^2)/2 + 8(5)] - [(1^3)/3 - 15(1^2)/2 + 8(1)] = -6.67 meters. Therefore, the displacement of the car is -6.67 meters.
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