How do you evaluate the limit of (2x^2 – 3x + 1)/(x – 2) as x approaches 2?To evaluate the limit of (2x^2 – 3x + 1)/(x – 2) as x approaches 2, factor the numerator. The numerator 2x^2 – 3x + 1 factors to (2x – 1)(x – 1). The expression becomes ((2x – 1)(x – 1))/(x – 2). As x approaches 2, the limit is determined by substituting x = 2 into the simplified expression, yielding a limit of 3.
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How do you divide polynomial functions and determine the quotient and the remainder using the polynomial long division method?
How do you divide polynomial functions and determine the quotient and the remainder using the polynomial long division method?To divide polynomial functions using polynomial long division, align terms by degree, divide the leading term of the dividend by the leading term of the divisor, multiply the entire divisor by this result, subtract from the dividend, and repeat the process with the new polynomial until the remainder is of lower degree than the divisor.
How do I determine the asymptotes and end behavior for the function f(x) = (3x^3 – 2x + 1) / (2x^2 – x – 3)?
How do I determine the asymptotes and end behavior for the function f(x) = (3x^3 – 2x + 1) / (2x^2 – x – 3)?To determine the asymptotes and end behavior for the function f(x) = (3x^3 – 2x + 1) / (2x^2 – x – 3), we analyze both vertical and horizontal asymptotes. Vertical asymptotes occur where the denominator equals zero, i.e., 2x^2 – x – 3 = 0. Solving this quadratic yields x = 3/2 and x = -1. Horizontal asymptotes depend on the degrees of the numerator and denominator. Here, the degree of the numerator (3) is greater than the degree of the denominator (2), indicating no horizontal asymptote. Instead, we find an oblique asymptote by performing polynomial long division, resulting in y = (3/2)x. The end behavior of the function follows the leading term of the polynomial division, meaning f(x) behaves like (3/2)x as x approaches ±∞.
What is the solution to the equation 2x + 5 = 15?
What is the solution to the equation 2x + 5 = 15?To solve the equation 2x + 5 = 15, first subtract 5 from both sides to get 2x = 10. Then, divide both sides by 2 to find x = 5. Thus, the solution to the equation is x = 5.
How do you find the derivative of a composite function using the chain rule?
How do you find the derivative of a composite function using the chain rule?To find the derivative of a composite function using the chain rule, you differentiate the outer function with respect to the inner function and then multiply by the derivative of the inner function. Mathematically, if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x). This method allows you to handle complex functions systematically.
How can the distributive property be used to simplify large multiplications in pre-algebra?
How can the distributive property be used to simplify large multiplications in pre-algebra?The distributive property allows you to break down large multiplications into smaller, more manageable parts. For example, to multiply 27 by 15, you can use the distributive property: 27 * 15 = 27 * (10 + 5) = (27 * 10) + (27 * 5). This simplifies the calculation to 270 + 135 = 405.
How do you find the derivative of a function using the power rule, and can you provide an example?
How do you find the derivative of a function using the power rule, and can you provide an example?The power rule states that if you have a function of the form f(x) = x^n, the derivative f'(x) = n*x^(n-1). For example, if f(x) = x^3, then f'(x) = 3*x^2.
If $f(x) = 2x^2 – 3x + 5$, find the value of x when $f(x)$ equals 12.
If f(x) = 2x^2 – 3x + 5, find the value of x when f(x) equals 12.To find the value of x when f(x) = 12, solve the equation 2x^2 – 3x + 5 = 12. Simplifying, we get 2x^2 – 3x – 7 = 0. Using the quadratic formula, x = (3 ± √(9 + 56))/4, yielding x = 2 or x = -7/2.
How do you derive the double angle formulas for sine, cosine, and tangent and apply them to solve complex trigonometric equations?
How do you derive the double angle formulas for sine, cosine, and tangent and apply them to solve complex trigonometric equations?To derive the double angle formulas for sine, cosine, and tangent, use the identities sin(2θ) = 2sin(θ)cos(θ), cos(2θ) = cos²(θ) – sin²(θ), and tan(2θ) = 2tan(θ)/(1-tan²(θ)). Apply these formulas by substituting them into trigonometric equations to simplify and solve for unknown angles.
How do you find the sine, cosine, and tangent of an angle in a right triangle using the Pythagorean Theorem?
How do you find the sine, cosine, and tangent of an angle in a right triangle using the Pythagorean Theorem?To find the sine, cosine, and tangent of an angle in a right triangle using the Pythagorean Theorem, first identify the lengths of the sides: opposite (a), adjacent (b), and hypotenuse (c). The Pythagorean Theorem states that a² + b² = c². Then, sine is a/c, cosine is b/c, and tangent is a/b.
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