Question

this question is designed to be answered without a calculator. let f(x) = 1 - 4x³. what is \\(\lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h}\\)? \\(\bigcirc\\) 12x² \\(\bigcirc\\) -12x² \\(\bigcirc\\) 12x² + 12x \\(\bigcirc\\) -12x² - 12x

Explanation:

Step1: Substitute \( f(x+h) \) and \( f(x) \)

Given \( f(x) = 1 - 4x^3 \), then \( f(x+h) = 1 - 4(x+h)^3 \). Expand \( (x+h)^3 \) using the formula \( (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \), so \( (x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3 \). Then \( f(x+h) = 1 - 4(x^3 + 3x^2h + 3xh^2 + h^3) = 1 - 4x^3 - 12x^2h - 12xh^2 - 4h^3 \). Now, \( f(x+h) - f(x) = [1 - 4x^3 - 12x^2h - 12xh^2 - 4h^3] - [1 - 4x^3] = -12x^2h - 12xh^2 - 4h^3 \).

Step2: Divide by \( h \) and take limit

Divide \( f(x+h) - f(x) \) by \( h \): \( \frac{f(x+h) - f(x)}{h} = \frac{-12x^2h - 12xh^2 - 4h^3}{h} = -12x^2 - 12xh - 4h^2 \). Now, take the limit as \( h \to 0 \): \( \lim_{h \to 0} (-12x^2 - 12xh - 4h^2) \). As \( h \to 0 \), the terms with \( h \) (i.e., \( -12xh \) and \( -4h^2 \)) approach 0. So the limit is \( -12x^2 \).

Answer:

\( -12x^2 \) (corresponding to the option: \( -12x^2 \))