Lesson 25 (10/28/26)
Graphing Functions - II
Warmup: Determine whether the following are odd or even
Even: \( f(-x) = f(x) \)
Y-axis symmetry
odd! \( f(-x) = -f(x) \)
Origin symmetry
Warmup: Determine whether the following are odd or even
Even: \( f(-x) = f(x) \)
Y-axis symmetry
odd! \( f(-x) = -f(x) \)
Origin symmetry
Domain
x-intercepts and y-intercepts: For x-intercepts set \( y=0 \), for y-intercepts set \( x=0 \).
Vertical Asymptotes: \( \lim_{x \to a^+} f(x) = \pm \infty \) OR \( \lim_{x \to a^-} f(x) = \pm \infty \).
End Behaviour: \( \lim_{x \to \pm \infty} f(x) \). This helps identify Horizontal Asymptotes, Slant Asymptotes, or No Asymptote.
\( f'(x) \): \( f(x) \uparrow \) if \( f'(x) > 0 \), \( f(x) \downarrow \) if \( f'(x) < 0 \). Points where \( f'(x) = 0 \) or DNE are critical points.
\( f''(x) \): \( f''(x) > 0 \) is concave up, \( f''(x) < 0 \) is concave down. Points of inflection.
Symmetry: Even function - y-axis symmetry, odd function - origin symmetry.
Domain: \( (-\infty, \infty) \)
Symmetry: \( f(-x) = e^{(-x)^2} = e^{x^2} = f(x) \). This is an even function and has y-axis symmetry.
Note: We can just graph on \( [0, \infty) \) and take the mirror image with respect to the y-axis.
X-intercept: Function value = 0 \(\implies\) \( e^{x^2} = 0 \). None, no x-intercept.
Y-intercept: \( y = f(0) = e^{0^2} = 1 \)
Vertical Asymptotes: \( f(x) \) has domain \( (-\infty, \infty) \) \(\implies\) No V.A.
\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} e^{-x^2} = 0 \] \[ \lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} e^{-x^2} = 0 \] Both limits imply that \( y=0 \) is the Horizontal Asymptote.
\[ f'(x) = \frac{d}{dx} [e^{-x^2}] = e^{-x^2} \cdot [-2x] = -(2e^{-x^2}) \cdot x \] Note: The term \( (2e^{-x^2}) > 0 \).
The sign analysis shows:
As indicated by the diagram, \( f(x) \) is increasing on \( (-\infty, 0) \) and decreasing on \( (0, \infty) \), resulting in a local max at \( x=0 \).
\(f''(x)\)!
\[ \frac{d}{dx} f'(x) = \frac{d}{dx} [-2e^{-x^2} \cdot x] \]
\[ = -2 \cdot \left( \frac{d}{dx} e^{-x^2} \right) \cdot x - 2e^{-x^2} \cdot \frac{d}{dx} x \]
\[ = (-2)(e^{-x^2} \cdot (-2x)) \cdot x - 2e^{-x^2} \cdot 1 \]
\[ = 2e^{-x^2} [2x^2 - 1] = 4e^{-x^2} [x^2 - 1/2] \]
Note: Under the expression \(e^{-x^2}\) there is an annotation indicating \(>0\).
\( f(x) = e^{-x^2} \)
Symmetry: \( f(x) = \sin x - x \)
\[ f(-x) = \sin(-x) - (-x) = -\sin x + x = -[\sin x - x] = -f(x) \]
Note: Odd function. Graph has origin symmetry.
X-intercept: \( f(x) = 0 \implies \sin x - x = 0 \implies \sin x = x \)
Note: \( x = 0 \) is the only such number.
Y-intercept: \( f(0) = \sin 0 - 0 = 0 \)
End Behaviour: Do not have to consider.
Vertical Asymptotes: NONE
Note: when \( \cos x = 1 \)
Observation: \( f' = \cos x - 1 < 0 \)
Observation: \( f' = \cos x - 1 < 0 \)
decreasing on \( (-2\pi, 0) \) & \( (0, 2\pi) \)
\[ f''(x) = \frac{d}{dx} f'(x) = \frac{d}{dx} [\cos x - 1] = -\sin x \]
Example: \( f'(x) = (x+4)(x-1)(x-5) \), sketch \( f(x) \)
Critical points: \( x=1 \), \( x=5 \), \( x=-4 \)
Increasing: \( (-4, 1) \), \( (5, \infty) \)
Decreasing: \( (-\infty, -4) \), \( (1, 5) \)
\[ f'' = \frac{d}{dx} f'(x) = \frac{d}{dx} \left[ (x+4)(x-1)(x-5) \right] \]
\[ = \left[ \frac{d}{dx} (x+4) \right] (x-1)(x-5) + (x+4) \left[ \frac{d}{dx} (x-1)(x-5) \right] \]
Note: Under the first bracket, the derivative is \( 1 \). Under the second bracket, the expansion is \( \left( \frac{d}{dx}(x-1) \right)(x-5) + (x-1) \frac{d}{dx}(x-5) \), where each derivative is \( 1 \).
\[ = (x-1)(x-5) + (x+4)(x-5) + (x+4)(x-1) \]
\[ = x^2 - 6x + 5 + x^2 - x - 20 + x^2 + 3x - 4 \]
\[ = 3x^2 - 4x - 19 \]
\[ 3x^2 - 4x - 19 = 0 \]
Note: This is a quadratic equation of the form \( ax^2 + bx + c = 0 \), which can be solved using the quadratic formula: \[ \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
The equation \( 3x^2 - 4x - 19 = 0 \) is solved using the quadratic formula:
\[ \frac{4 \pm \sqrt{16 + 4(3)(-19)}}{6} \]
The roots are calculated as:
\[ x_1 = \frac{2 + \sqrt{61}}{3}, \quad x_2 = \frac{2 - \sqrt{61}}{3} \]