Graph Concept: A secant line connects two points on a curve. The handwritten notes show points (0,0), (1,1), (2,4), and (3,9) on the parabola y = x², with secant lines drawn between various pairs of points.
Calculate the slopes of secant lines through various points:
Solution:
As \(x \to 0\), the slope approaches 0, giving us the slope of the tangent line at \(x = 0\).
Derivative of a function: The slope of the tangent line at \(x = a\)
The derivative at \(x = a\) is the limit of the slope of secant lines as the second point approaches \(a\):
\[f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}\]This limit exists if it is finite. Otherwise, we say the derivative does not exist at \(x = a\).
Alternative Definition: Instead of using a secant line through \((a, f(a))\) and \((x, f(x))\), we can use \((a, f(a))\) and \((a+h, f(a+h))\):
\[f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\]Use the definition to find \(f'(2)\).
Solution:
\[f'(2) = \lim_{x \to 2} \frac{f(x) - f(2)}{x - 2} = \lim_{x \to 2} \frac{x^2 - 4}{x - 2}\]Factor the numerator: \(x^2 - 4 = (x-2)(x+2)\)
\[f'(2) = \lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2} (x+2) = 4\]Find the equation of the tangent line to \(y = f(x) = x^2\) at the point (2,4).
Solution:
We need: Point = (2,4) and Slope = \(f'(2) = 4\)
Using point-slope form: \(y - y_0 = m(x - x_0)\)
\[y - 4 = 4(x - 2)\] \[y = 4x - 8 + 4 = 4x - 4\]Find a formula for \(f'(a)\) when \(f(x) = x^2\).
Solution:
\[f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} = \lim_{x \to a} \frac{x^2 - a^2}{x - a}\]Factor: \(x^2 - a^2 = (x-a)(x+a)\)
\[f'(a) = \lim_{x \to a} \frac{(x-a)(x+a)}{x-a} = \lim_{x \to a} (x+a) = 2a\]Therefore, \(f'(x) = 2x\)
If \(f(x) = x^2 - 7x\), find \(f'(3)\).
Solution:
\[f'(3) = \lim_{h \to 0} \frac{f(3+h) - f(3)}{h}\]Calculate \(f(3) = 9 - 21 = -12\)
Calculate \(f(3+h) = (3+h)^2 - 7(3+h) = 9 + 6h + h^2 - 21 - 7h = h^2 - h - 12\)
\[f'(3) = \lim_{h \to 0} \frac{h^2 - h - 12 - (-12)}{h} = \lim_{h \to 0} \frac{h^2 - h}{h} = \lim_{h \to 0} (h - 1) = -1\]Find \(f'(4)\) if \(f(x) = \sqrt{x}\).
Solution:
\[f'(4) = \lim_{h \to 0} \frac{f(4+h) - f(4)}{h} = \lim_{h \to 0} \frac{\sqrt{4+h} - 2}{h}\]Rationalize the numerator by multiplying by \(\frac{\sqrt{4+h} + 2}{\sqrt{4+h} + 2}\):
\[f'(4) = \lim_{h \to 0} \frac{(\sqrt{4+h} - 2)(\sqrt{4+h} + 2)}{h(\sqrt{4+h} + 2)} = \lim_{h \to 0} \frac{4+h - 4}{h(\sqrt{4+h} + 2)}\] \[= \lim_{h \to 0} \frac{h}{h(\sqrt{4+h} + 2)} = \lim_{h \to 0} \frac{1}{\sqrt{4+h} + 2} = \frac{1}{4}\]Suppose \(f'(5) = \lim_{h \to 0} \frac{(5+h)^3 - 125}{h}\). What is the function \(f(x)\) and the value \(a\)?
Solution:
Comparing with the definition \(f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\):
To compute \(f'(5)\):
\[(5+h)^3 = 125 + 75h + 15h^2 + h^3\] \[f'(5) = \lim_{h \to 0} \frac{75h + 15h^2 + h^3}{h} = \lim_{h \to 0} (75 + 15h + h^2) = 75\]Key Concept: The derivative gives us the instantaneous rate of change of a function, which geometrically represents the slope of the tangent line to the curve at any given point.