Let \(f(x) = 3x^2 - 8\). Find \(f'(-2)\) and \(f'(1)\).
Using the definition:
\[f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}\]For \(f'(-2)\):
\[f'(-2) = \lim_{h \to 0} \frac{f(-2+h) - f(-2)}{h}\]Since this is a linear function in the numerator after simplification:
\[f'(-2) = 6(-2) = -12\]For \(f'(1)\):
\[f'(1) = 6(1) = 6\]General pattern: For \(f(x) = 3x^2 - 8\), we get \(f'(x) = 6x\)
Derivative Function: Instead of finding the derivative at a specific point, we can find a formula that gives the derivative at any point \(x\):
\[f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]This creates a new function \(f'(x)\) where each input \(x\) produces the slope of the tangent line at that point.
Find \(f'(x)\) if \(f(x) = 3x + 8\).
Solution:
\[f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\] \[= \lim_{h \to 0} \frac{[3(a+h) + 8] - [3a + 8]}{h}\] \[= \lim_{h \to 0} \frac{3a + 3h + 8 - 3a - 8}{h} = \lim_{h \to 0} \frac{3h}{h} = 3\]Therefore, \(f'(x) = 3\) for all \(x\).
Interpretation: The slope is constant at 3 everywhere (it's a line!).
Find \(f'(x)\) if \(f(x) = \sqrt{3x - 8}\).
Solution:
\[f'(x) = \lim_{h \to 0} \frac{\sqrt{3(x+h) - 8} - \sqrt{3x - 8}}{h}\]Rationalize by multiplying by \(\frac{\sqrt{3(x+h) - 8} + \sqrt{3x - 8}}{\sqrt{3(x+h) - 8} + \sqrt{3x - 8}}\):
\[= \lim_{h \to 0} \frac{[3(x+h) - 8] - [3x - 8]}{h[\sqrt{3(x+h) - 8} + \sqrt{3x - 8}]}\] \[= \lim_{h \to 0} \frac{3h}{h[\sqrt{3(x+h) - 8} + \sqrt{3x - 8}]} = \frac{3}{2\sqrt{3x - 8}}\]Domain: \(x > \frac{8}{3}\) (where \(f(x)\) is defined)
\(f'(x)\) does not exist at \(x = \frac{8}{3}\)
The derivative \(f'(a)\) does not exist when:
Consider the piecewise function:
\[f(x) = \begin{cases} 3x^2 + 8 & \text{if } x \geq 0 \\ x + 5 & \text{if } x < 0 \end{cases}\]Find \(f'(0)\).
Solution:
Check if \(f(0)\) is defined: \(f(0) = 3(0)^2 + 8 = 8\) ✓
Right-hand limit:
\[\lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{3h^2 + 8 - 8}{h} = \lim_{h \to 0^+} 3h = 0\]Left-hand limit:
\[\lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{h + 5 - 8}{h} = \lim_{h \to 0^-} \frac{h - 3}{h} = -\infty\]Since the left-hand limit does not exist (approaches \(-\infty\)), \(f'(0)\) does not exist.
Find \(f'(0)\) for \(f(x) = |x|\).
Solution:
\(f(0) = 0\), so \(f\) is defined at 0.
Right-hand derivative:
\[\lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{|h| - 0}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1\]Left-hand derivative:
\[\lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{|h| - 0}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1\]Since \(1 \neq -1\), the limit does not exist, so \(f'(0)\) does not exist.
Geometric interpretation: There's a sharp corner at \(x = 0\).
Graph Analysis: The handwritten notes show two graphs side by side - one showing f(x) and another showing f'(x). Key relationships are illustrated: where f(x) is increasing, f'(x) is positive; where f(x) is decreasing, f'(x) is negative; where f(x) has horizontal tangents, f'(x) = 0.
Key Relationships:
Given the graph of \(f(x)\), sketch \(f'(x)\).
Graph Features: The original function shows various slopes: negative slope (-1) in one region, zero slope at a minimum point, positive slope (4) in another region, and zero slope again at a maximum, followed by negative slope (-1).
Analysis:
Given the graph of the derivative \(f'(x)\), sketch the original function \(f(x)\).
Analysis: