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3 years ago

Solve for |4x+3| < 9

1 answer:

7 0

Under the absolute value could be positive or negative number, but absolute value make them both positive.
For positive number we can write is (4x+3) and for negative -(4x+3)
4x+3<9
4x< 6
x< 3/2 (-∞, 3/2)

-(4x+3)<9
-4x-12<9
-4x<21
x>-21/4 (-21/4, +∞)

(-∞, 3/2)∩ (-21/4, +∞) =(-21/4, 3/2) it could be written as (-5.25, 1.5)

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$\displaystyle y'(1, \frac{3}{2}) = -3$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Coordinates (x, y)
Functions
Function Notation
Terms/Coefficients
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

Step 1: Define

$\displaystyle y = \frac{3}{2x^2}$

$\displaystyle \text{Point} \ (1, \frac{3}{2})$

Step 2: Differentiate

[Function] Rewrite [Exponential Rule - Rewrite]: $\displaystyle y = \frac{3}{2}x^{-2}$
Basic Power Rule: $\displaystyle y' = -2 \cdot \frac{3}{2}x^{-2 - 1}$
Simplify: $\displaystyle y' = -3x^{-3}$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle y' = \frac{-3}{x^3}$

Step 3: Solve

Substitute in coordinate [Derivative]: $\displaystyle y'(1, \frac{3}{2}) = \frac{-3}{1^3}$
Evaluate exponents: $\displaystyle y'(1, \frac{3}{2}) = \frac{-3}{1}$
Divide: $\displaystyle y'(1, \frac{3}{2}) = -3$