Lesson 10 (3.2)

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Lesson 10 (3.2)

Warmup Example: Let \( f(x) = 3x^2 + 8 \), find \( f'(-2) \) and \( f'(1) \)

\[ f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \]

Finding \( f'(-2) \):

\[ \begin{aligned} f'(-2) &= \lim_{h \to 0} \frac{f(-2+h) - f(-2)}{h} \\ &= \lim_{h \to 0} \frac{(3(-2+h)^2 + 8) - (3(-2)^2 + 8)}{h} \\ &= \lim_{h \to 0} \frac{3(-2)^2 + 3[-2 \times 2h] + h^2 - 3(-2)^2}{h} \\ &= \lim_{h \to 0} \frac{6h(-2) + h^2}{h} \\ &= 6(-2) = -12 \end{aligned} \]

Finding \( f'(1) \):

\[ \begin{aligned} f'(1) &= \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} \\ &= \lim_{h \to 0} \frac{(3(1+h)^2 + 8) - (3(1)^2 + 8)}{h} \\ &= \lim_{h \to 0} \frac{3(1)^2 + 3[2 \times 1h] + h^2 - 3(1)^2}{h} \\ &= \lim_{h \to 0} \frac{6h(1) + h^2}{h} \\ &= 6(1) = 6 \end{aligned} \]

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What if I want to compute at a generic point \( x \)?

\[ \begin{aligned} f'(x) &= \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\ &= \lim_{h \to 0} \frac{(3(x+h)^2 + 8) - (3(x)^2 + 8)}{h} \\ &= \lim_{h \to 0} \frac{3(x)^2 + 3[2x \cdot h] + h^2 - 3(x)^2}{h} \\ &= \lim_{h \to 0} \frac{6h(x) + h^2}{h} = 6x \end{aligned} \]

\( f'(x) \) is a function with input \( x \)

output = derivative at \( x \)
= slope of tangent line at \( (x, f(x)) \)

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Derivative as a function

A 2D Cartesian coordinate plane illustrating the geometric interpretation of the derivative. A continuous curve labeled y = f(x) traverses the plane. A vertical dashed line extends from a marked point x on the horizontal axis up to the curve. At this point on the curve, a straight tangent line is drawn, annotated with the text 'tangent line with slope f'(x)' to indicate that the derivative represents the slope of the tangent line at x.

Visual Description: A 2D Cartesian coordinate plane illustrating the geometric interpretation of the derivative. A continuous curve labeled y = f(x) traverses the plane. A vertical dashed line extends from a marked point x on the horizontal axis up to the curve. At this point on the curve, a straight tangent line is drawn, annotated with the text 'tangent line with slope f'(x)' to indicate that the derivative represents the slope of the tangent line at x.

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

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Example: \( f(x) = \sqrt{3x-8} \), find \( f'(x) \).

Domain: \( 3x - 8 \ge 0 \), \( x \ge 8/3 \)
\( [8/3, \infty) \)

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] \[ = \lim_{h \to 0} \frac{\sqrt{3(x+h) - 8} - \sqrt{3x - 8}}{h} \] \[ = \lim_{h \to 0} \frac{(\sqrt{3x+3h-8} - \sqrt{3x-8})(\sqrt{3x+3h-8} + \sqrt{3x-8})}{h(\sqrt{3x+3h-8} + \sqrt{3x-8})} \] \[ = \lim_{h \to 0} \frac{(3x+3h-8) - (3x-8)}{h(\sqrt{3x+3h-8} + \sqrt{3x-8})} = \lim_{h \to 0} \frac{3h}{h(\sqrt{3x+3h-8} + \sqrt{3x-8})} \] \[ = \frac{3}{2\sqrt{3x-8}} \]

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\[ f(x) = \sqrt{3x - 8} \qquad D: \left[\frac{8}{3}, \infty\right) \]

\[ f'(x) = \frac{3}{2\sqrt{3x - 8}} \qquad \text{w/ } D: \left(\frac{8}{3}, \infty\right) \]

\( f'\left(\frac{8}{3}\right) \) is not defined!!

Q: What happens to graph as \( x \to \frac{8}{3} \)

\[ f'(x) = \frac{3}{2\sqrt{3x - 8}} \]

\[ \lim_{x \to \frac{8}{3}^+} f'(x) = \lim_{x \to \frac{8}{3}^+} \frac{3}{2\sqrt{3x - 8}} = \frac{3}{\substack{\text{small} \\ +\text{ve}}} = +\infty \]

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\[ f(x) = \sqrt{3x-8} \quad , \quad f'(x) = \frac{3}{2\sqrt{3x-8}} \]

Cartesian coordinate graph of the function f(x) = sqrt(3x-8) plotted in blue, starting at the point (8/3, 0) on the x-axis and increasing concaved downwards into the first quadrant. A vertical purple line is drawn at x = 8/3, indicating a vertical tangent at the starting point where the slope approaches infinity. A secondary tangent line is shown touching the curve at a higher x value.

Visual Description: Cartesian coordinate graph of the function f(x) = sqrt(3x-8) plotted in blue, starting at the point (8/3, 0) on the x-axis and increasing concaved downwards into the first quadrant. A vertical purple line is drawn at x = 8/3, indicating a vertical tangent at the starting point where the slope approaches infinity. A secondary tangent line is shown touching the curve at a higher x value.

\[ \lim_{x \to 8/3^+} f'(x) = +\infty \implies f'(8/3) \text{ DNE} \]

at \( 8/3 \), \( f(x) \) has a vertical tangent

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Example:

\[ f(x) = \begin{cases} 3x^2 + 8 & x \le 0 \\ x + 5 & x > 0. \end{cases} \]

Graph of the piecewise function f(x) on a Cartesian coordinate plane. For x <= 0, the graph shows a parabolic curve y = 3x^2 + 8 in the second quadrant decreasing down to a solid closed circle at the y-intercept (0, 8). For x > 0, the graph is a straight line y = x + 5 starting from an open circle at (0, 5) on the y-axis and extending upwards with a slope of 1 into the first quadrant.

Visual Description: Graph of the piecewise function f(x) on a Cartesian coordinate plane. For x <= 0, the graph shows a parabolic curve y = 3x^2 + 8 in the second quadrant decreasing down to a solid closed circle at the y-intercept (0, 8). For x > 0, the graph is a straight line y = x + 5 starting from an open circle at (0, 5) on the y-axis and extending upwards with a slope of 1 into the first quadrant.

Find \( f'(x) \)

\( x < 0 \leadsto \)

\[ \begin{aligned} f(x) &= 3x^2 + 8 \\ f'(x) &= 6x \end{aligned} \]

\( f'(x) \) exists for all \( x < 0 \)

\( x > 0 \leadsto \)

\[ \begin{aligned} f(x) &= x + 5 \\ f'(x) &= 1 \end{aligned} \]

\( f'(x) \) exists for all \( x > 0 \)

at \( x = 0 \)?

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\[ f(x) = \begin{cases} 3x^2 + 8 & x \le 0 \\ x + 5 & x > 0 \end{cases} \]

\[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \]

Left-hand limit:

\[ \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{3h^2 + 8 - 8}{h} = 0 \]

Right-hand limit:

\[ \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h + 5 - 8}{h} = \lim_{h \to 0^+} \left(1 - \frac{3}{h}\right) = -\infty \]

\[ f'(0) \text{ DNE} \]

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Example

\[ f(x) = \begin{cases} 3x^2 + 8 & x \le 0 \\ x + 8 & x > 0 \end{cases} \]

Cartesian coordinate graph of the piecewise function f(x). For x ≤ 0, the graph is a parabolic curve y = 3x^2 + 8 decreasing from the second quadrant to a minimum at the y-intercept (0, 8), indicated by a blue point. For x > 0, the graph proceeds into the first quadrant as a straight line y = x + 8 with slope 1, forming a corner at the point (0, 8).

Visual Description: Cartesian coordinate graph of the piecewise function f(x). For x ≤ 0, the graph is a parabolic curve y = 3x^2 + 8 decreasing from the second quadrant to a minimum at the y-intercept (0, 8), indicated by a blue point. For x > 0, the graph proceeds into the first quadrant as a straight line y = x + 8 with slope 1, forming a corner at the point (0, 8).

\( f'(0) \text{ still DNE} \)

\[ \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{3h^2 + 8 - 8}{h} = 0 \]

\[ \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h + 8 - 8}{h} = 1 \]

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Example: \( f(x) = |x| \), find \( f'(x) \)

\[ f(x) = \begin{cases} x & x \ge 0 \\ -x & x < 0 \end{cases} \]

Graph of the absolute value function f(x) = |x| plotted on a Cartesian coordinate system. It shows a continuous V-shaped graph with its vertex at the origin (0,0), extending into the second quadrant with slope -1 for x < 0 and into the first quadrant with slope +1 for x > 0.

Visual Description: Graph of the absolute value function f(x) = |x| plotted on a Cartesian coordinate system. It shows a continuous V-shaped graph with its vertex at the origin (0,0), extending into the second quadrant with slope -1 for x < 0 and into the first quadrant with slope +1 for x > 0.

\[ x < 0 \implies f'(x) = -1 \] \[ x > 0 \implies f'(x) = 1 \] \[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \] \[ = \lim_{h \to 0} \frac{|h|}{h} \]

Taking one-sided limits:

As \( h \to 0^+ \implies 1 \)
As \( h \to 0^- \implies -1 \)

\[ f'(0) \text{ DNE} \]

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Summary:

\(f(x)\) is NOT differentiable at \(x = a\) \(\Big\}\) \(f'(a)\) DNE

\(f\) is not continuous at \(x = a\) \(\Big\}\) hole, jump disc., V.A
\(f\) has a vertical tangent at \(x = a\)
\(f\) has a corner or a cusp

Visual Description: Hand-drawn sketch illustrating a corner on a graph, depicted as a sharp V-shaped point where two straight line segments meet.

Visual Description: Hand-drawn sketch illustrating a cusp on a graph, showing two curved line segments tapering into a sharp, pointed vertex.

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Graphing \( f'(x) \) from \( f(x) \)

\( f(x) = 3x^2 + 8 \)

Visual Description: Graph of a parabola representing f(x) = 3x^2 + 8 centered on the vertical y-axis above the horizontal x-axis. Three tangent lines illustrate the sign of the derivative: a downward-sloping tangent line on the decreasing left branch labeled f' < 0, a horizontal tangent line at the minimum vertex labeled f' = 0, and an upward-sloping tangent line on the increasing right branch labeled f' > 0.

\( f'(x) = 6x \)

Visual Description: Cartesian coordinate graph of the linear derivative function f'(x) = 6x, plotted as a straight line passing through the origin (0, 0) with a positive slope, moving from the third quadrant through to the first quadrant.

\( f' < 0 \implies f \text{ is decreasing} \)

\( f' > 0 \implies f \text{ is increasing.} \)

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Example: Use the graph of \( f(x) \) to sketch \( f'(x) \)

Graph of a piecewise linear function f(x) on the interval [-8, 8]. From x = -8 to x = -4, a straight line segment goes from (-8, 4) down to (-4, 0). From x = -4 to x = 0, the line segment rises from (-4, 0) to an open circle at (0, 4). From x = 0 to x = 4, a horizontal line segment connects (0, 2) to (4, 2). From x = 4 to x = 8, a line segment starts with an open circle at (4, -2) and rises to (8, 0).

Visual Description: Graph of a piecewise linear function f(x) on the interval [-8, 8]. From x = -8 to x = -4, a straight line segment goes from (-8, 4) down to (-4, 0). From x = -4 to x = 0, the line segment rises from (-4, 0) to an open circle at (0, 4). From x = 0 to x = 4, a horizontal line segment connects (0, 2) to (4, 2). From x = 4 to x = 8, a line segment starts with an open circle at (4, -2) and rises to (8, 0).

\( -8 < x < -4 \leadsto \text{slope} = -1 \)
\( -4 < x < 0 \leadsto \text{slope} = 1 \)
\( 0 < x < 4 \leadsto \text{slope} = 0 \)
\( 4 < x < 8 \leadsto \text{slope} = 1/2 \)

Graph of the derivative function f'(x) plotted on the same domain [-8, 8]. It consists of horizontal line segments corresponding to the slopes: y = -1 on the interval (-8, -4) with open endpoints; y = 1 on the interval (-4, 0) with open endpoints; y = 0 on the interval (0, 4) with open endpoints along the x-axis; and y = 1/2 on the interval (4, 8) with an open circle at x = 4 and a closed dot at x = 8.

Visual Description: Graph of the derivative function f'(x) plotted on the same domain [-8, 8]. It consists of horizontal line segments corresponding to the slopes: y = -1 on the interval (-8, -4) with open endpoints; y = 1 on the interval (-4, 0) with open endpoints; y = 0 on the interval (0, 4) with open endpoints along the x-axis; and y = 1/2 on the interval (4, 8) with an open circle at x = 4 and a closed dot at x = 8.

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Graphing \( f(x) \) from \( f'(x) \)

Graph of the derivative function f'(x) on the interval [-4, 4]. The graph consists of piecewise linear segments: a linear segment starting from a solid dot at (-4, -2), crossing the x-axis at (-2, 0), and terminating at an open circle at (0, 2); a horizontal segment along the x-axis from (0, 0) to (2, 0) with solid endpoints; and a downward-sloping linear segment from an open circle at (2, 2) down to a solid dot at the x-intercept (4, 0). Dashed guide lines mark coordinates at (-4, -2) and connect the open circles at y = 2.

Visual Description: Graph of the derivative function f'(x) on the interval [-4, 4]. The graph consists of piecewise linear segments: a linear segment starting from a solid dot at (-4, -2), crossing the x-axis at (-2, 0), and terminating at an open circle at (0, 2); a horizontal segment along the x-axis from (0, 0) to (2, 0) with solid endpoints; and a downward-sloping linear segment from an open circle at (2, 2) down to a solid dot at the x-intercept (4, 0). Dashed guide lines mark coordinates at (-4, -2) and connect the open circles at y = 2.

Graph of the function f(x) corresponding to the derivative f'(x). On [-4, 0), the curve begins at a solid dot at x = -4, decreases to a local minimum with a horizontal tangent line indicated at x = -2, and then increases up to a peak at x = 0. At x = 0, the function jumps down to a horizontal line segment on (0, 2) with an open circle at x = 2. At x = 2, it jumps up to an increasing, concave-down curve that levels off to a horizontal tangent line at x = 4, marked by a dashed vertical guideline down to x = 4.

Visual Description: Graph of the function f(x) corresponding to the derivative f'(x). On [-4, 0), the curve begins at a solid dot at x = -4, decreases to a local minimum with a horizontal tangent line indicated at x = -2, and then increases up to a peak at x = 0. At x = 0, the function jumps down to a horizontal line segment on (0, 2) with an open circle at x = 2. At x = 2, it jumps up to an increasing, concave-down curve that levels off to a horizontal tangent line at x = 4, marked by a dashed vertical guideline down to x = 4.

\( -4 \le x \le -2 \rightsquigarrow f'(x) < 0 \Rightarrow f \text{ is decreasing} \)

\( f'(-2) = 0 \Rightarrow f \text{ has horizontal tangent at } x = -2 \)

\( -2 < x < 0 \rightsquigarrow f'(x) > 0 \Rightarrow f \text{ is increasing} \)

\( 0 < x < 2 \rightsquigarrow f'(x) = 0 \Rightarrow f \text{ is constant} \)

\( 2 < x < 4 \rightsquigarrow f'(x) > 0 \Rightarrow f \text{ is increasing} \)

\( f'(4) = 0 \Rightarrow f \text{ has horizontal tangent at } x = 4 \)