The proof of the derivative of the quadratic functions is presented using the limit definition of the derivative.
Proof of the Derivative of the Quadratic Function
The definition of the derivative \( f' \) of a function \( f \) is given by the limit:
\[ f'(x) = \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} \]Let \( f \) be a quadratic function of the form \( f(x) = a x^2 + bx + c \). Write the derivative of \( f \) as follows:
\[ f'(x) = \lim_{h \to 0} \dfrac{a(x+h)^2 + b(x+h) + c - (ax^2 + bx + c)}{h} \]Expand the terms \( a(x+h)^2 \) and \( b(x+h) \) in the numerator:
\[ f'(x) = \lim_{h \to 0} \dfrac{ax^2 + ah^2 + 2axh + bx + bh + c - ax^2 - bx - c}{h} \]Simplify the numerator:
\[ f'(x) = \lim_{h \to 0} \dfrac{ah^2 + 2axh + bh}{h} \]Divide numerator and denominator by \( h \):
\[ f'(x) = \lim_{h \to 0} (ah + 2ax + b) \]Evaluate the limit to obtain the derivative of the quadratic function:
Examples with Solutions
Part A & Part B: Practice Problems and Solutions
Part A: Find the derivatives of the quadratic functions given by:
- \( f(x) = 4x^2 - x + 1 \)
- \( g(x) = -x^2 - 1 \)
- \( h(x) = 0.1x^2 - \dfrac{x}{2} - 100 \)
- \( f(x) = -\dfrac{3x^2}{7} - 0.2x + 7 \)
Part B: Let \( f(x) = ax^2 + bx + c \). Find \( f'(2) \) given that \( f(2) = 3 \), \( f'(0) = 1 \), and \( f'(-1) = 2 \).
Solutions:
Part A Solutions:
- \( f'(x) = 8x - 1 \)
- \( g'(x) = -2x \)
- \( h'(x) = 0.2x - \dfrac{1}{2} \)
- \( f'(x) = -\dfrac{6x}{7} - 0.2 \)
Part B Solution:
The given function \( f \) is a quadratic function, hence:
\[ f'(x) = 2ax + b \]Given \( f'(0) = 1 \), substitute to obtain the equation: \( 2a(0) + b = 1 \implies b = 1 \)
Given \( f'(-1) = 2 \), substitute to obtain the equation: \( 2a(-1) + 1 = 2 \implies -2a = 1 \implies a = -\dfrac{1}{2} \)
Given \( f(2) = 3 \), substitute to obtain the equation: \( f(2) = -\dfrac{1}{2}(2)^2 + (1)(2) + c = 3 \implies -2 + 2 + c = 3 \implies c = 3 \)
Finally, find \( f'(2) \):
\[ f'(2) = 2\left(-\dfrac{1}{2}\right)(2) + 1 = -2 + 1 = -1 \]