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Calculus I Homework HelpQ1: Differentiate the function f(x) = 3x² + 2x - 5.
See AnswerThe derivative of f(x) is f'(x) = 6x + 2.
Q2: Find the indefinite integral of the function f(x) = 4x³ - 2x² + x.
See AnswerThe indefinite integral is F(x) = x⁴ - (2/3)x³ + (1/2)x² + C, where C is the constant of integration.
Q3: Evaluate the limit: lim (x → 2) (x² - 4)/(x - 2).
See AnswerFactor the numerator: (x - 2)(x + 2). Cancel the (x - 2) terms, and the limit becomes lim (x → 2) (x + 2) = 4.
Q4: Use the definition of the derivative to find f'(x) for f(x) = x².
See AnswerUsing the limit definition of the derivative: f'(x) = lim (h → 0) [(x+h)² - x²]/h = 2x.
Q5: Find the area under the curve of f(x) = x² from x = 0 to x = 2.
See AnswerThe area is the definite integral of x² from 0 to 2: ∫[0,2] x² dx = (1/3)x³ | from 0 to 2 = 8/3.
Q6: Solve the differential equation dy/dx = 3x² with the initial condition y(0) = 4.
See AnswerIntegrating both sides: y(x) = x³ + C. Using the initial condition y(0) = 4, we find C = 4. So, y(x) = x³ + 4.
Q7: Find the critical points of the function f(x) = x³ - 6x² + 9x + 1.
See AnswerFirst, find the derivative: f'(x) = 3x² - 12x + 9. Set f'(x) = 0 and solve for x: 3x² - 12x + 9 = 0, giving critical points at x = 1.
Q8: Find the equation of the tangent line to the curve y = x² at the point (1, 1).
See AnswerThe slope of the tangent line is given by the derivative, which is 2x. At x = 1, the slope is 2. The equation of the tangent line is y - 1 = 2(x - 1), or y = 2x - 1.
Q9: Find the maximum and minimum values of the function f(x) = -x² + 4x + 1 on the interval [0, 3].
See AnswerFind the derivative: f'(x) = -2x + 4. Set f'(x) = 0 and solve for x: x = 2. Evaluate f(0), f(2), and f(3). The maximum value is f(2) = 5, and the minimum value is f(3) = 1.
Q10: Find the length of the curve y = sqrt(x) from x = 1 to x = 4.
See AnswerThe length of the curve is found using the formula L = ∫[a,b] sqrt(1 + (dy/dx)²) dx. Here, dy/dx = 1/(2sqrt(x)). The final integral evaluates to approximately 2.478.
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