3-1-5 Numerical Analysis

The first matrix has first row a, c and second row b, d. The inverse has scalar numerator 1 and denominator a d minus b c, followed by rows d, minus c and minus b, a. The product A times its inverse is I, shown with rows 1, 0 and 0, 1.

The original simultaneous equations are 3x + 2y = 31 and 4x + 3y = 44. Both source steps, matrix rows, inverse entries, products and final values x = 5, y = 8 are retained.

Examples: Calculate a solution of the following simultaneous linear equations. 3x + 2y = 31 4x + 3y = 44 1) Represent the simultaneous linear equations in a matrix. 2) Multiply both sides by the inverse matrix to calculate x and y.

The original quadratic equation, its two factors, both zero-factor equations and the final alternatives minus 2.5 and 3 are transcribed in their source order.

Examples: Calculate a solution of the following quadratic equation. 2x² − x − 15 = 0 1) Factorize the quadratic equation. 2x² − x − 15 = (2x + 5) (x − 3) 2) Calculate x so that one of the terms becomes 0 from the fact that the equation equals to 0 by multiplication. 2x + 5 = 0 ··· x = −2.5 x − 3 = 0 ··· x = 3 x = −2.5 or x = 3

The original paragraph displays a = b raised to x, x = log base b of a, and the common-logarithm example log base 10 of 2 approximately 0.3010. The lower base labels and exponent positions are retained.

A logarithm refers to the x in an equation, such as a = b ^x, and is represented as x = log b a (which is read as “x is the log base b of a”). Logarithms of base 10 are particularly called common logarithms, and also the base number 10 is often omitted (e.g., log x) (log102 ≈ 0.3010.) The following laws apply to logarithm:

The original Law 1, Law 2 and Law 3 rows retain their bases, exponent, division sign and printed restriction n > 0, n not equal to 1.

[Law 1] log b a ⁿ = n × log b a [Law 2] log b (x × y) = log b x + log b y [Law 3] log b a = log n a ÷ log n b (n > 0, n ≠ 1)

The left box reads Function: x raised to n. The right reads Derivative: n times x raised to n minus 1. The upper arrow points right and is labelled Differentiation, with f(x) and f-prime(x). The lower arrow points left and is labelled Integration, with the integral of f(x) dx and f(x).

The original example evaluates the derivative of x cubed minus 6x minus 4 at x = 2. The expanded source differentiation row includes its printed zero-coefficient term and exponent zero minus one; no source term is omitted.

Examples: Calculate the slope of the tangent to the following function f(x) when x = 2. f(x) = x³ − 6x − 4 1) Calculate the derivative f’(x) of the function f(x). f’(x) = 1 × 3 × x³⁻¹ − 6 × 1 × x¹⁻¹ − 4 × 0 × x⁰⁻¹ = 3x² − 6 2) Calculate the slope of the tangent when x = 2. f’(2) = 3 × 2² − 6 = 6

The source function is 6x squared minus 5. The original indefinite integral retains the constant C. The definite-integral bounds are 1 below and 2 above the integral sign. The complete substitution and source result 3 remain intact.

Examples: Calculate the area of the interval [1, 2] (x = 1, x = 2) which is enclosed by the function f(x) and the x axis. f(x) = 6x² − 5 1) Calculate F(x) to get ∫ f(x)dx = F(x) + C. (indefinite integral) (C is the constant of integration.) F(x) = (6 ÷ 3) × x^{2 + 1} − (5 ÷ 1) × x^{0 + 1} + C = 2x³ − 5x + C 2) Calculate the area of the interval [1, 2]. (definite integral).

The original rising curve is labelled y = f(x), with axes x and y and a point labelled Solution. The three numbered interval rows mark x1, x2 and x3 with lower indices. The source callout states f(x1) > 0 and f(x2) < 0 in its second and third steps, and repeats until the solution is found.

The graph has axes x and y, 0, and a dashed curve labelled f(x). Filled points mark known data and the open point marks unknown data. The callout says to derive the polynomial f(x) from known data to find the unknown data.