Question

Evaluate (if possible) the function at the given value(s) of the independent variable. Simplify the results. f(x)

Answer 1

Answer:

$\frac{f(x + \triangle x) - f(x)}{\triangle x} = 3x^2+ 3x \cdot \triangle x + (\triangle x)^2$

Step-by-step explanation:

Given

$f(x) = x^3$

Required

Evaluate

$\frac{f(x + \triangle x) - f(x)}{\triangle x}$

$\frac{f(x + \triangle x) - f(x)}{\triangle x}$ becomes

$\frac{f(x + \triangle x) - f(x)}{\triangle x} = \frac{(x + \triangle x)^3 - x^3}{\triangle x}$

Expand

$\frac{f(x + \triangle x) - f(x)}{\triangle x} = \frac{x^3 + 3x^2 \cdot \triangle x+ 3x \cdot (\triangle x)^2 + (\triangle x)^3 - x^3}{\triangle x}$

Collect like terms

$\frac{f(x + \triangle x) - f(x)}{\triangle x} = \frac{x^3 - x^3+ 3x^2 \cdot \triangle x+ 3x \cdot (\triangle x)^2 + (\triangle x)^3 }{\triangle x}$

$\frac{f(x + \triangle x) - f(x)}{\triangle x} = \frac{3x^2 \cdot \triangle x+ 3x \cdot (\triangle x)^2 + (\triangle x)^3 }{\triangle x}$

Factorize

$\frac{f(x + \triangle x) - f(x)}{\triangle x} = \frac{\triangle x(3x^2+ 3x \cdot \triangle x + (\triangle x)^2) }{\triangle x}$

$\frac{f(x + \triangle x) - f(x)}{\triangle x} = 3x^2+ 3x \cdot \triangle x + (\triangle x)^2$