Precalculus Introduction, Basic Overview, Graphing Parent Functions, Transformations, Domain & Range

The cube root function y=³√x is similar to the square root function but is symmetric about the origin, extending into negative x and y values. Both its domain and range are (-∞, ∞).

The video introduces pre-calculus functions, focusing on graphing, domain, and range. It begins with the linear function y=x, a straight line passing through the origin. The domain is (-∞, ∞) as x can be any real number, and the range is also (-∞, ∞) as y can be any real number.

Next is the quadratic function y=x², which forms an upward-opening parabola. Its domain is (-∞, ∞), but its range is [0, ∞) because the lowest y-value is 0, and it never goes below the x-axis.

The cubic function y=x³ is an increasing function with an 'S' shape. Both its domain and range are (-∞, ∞), meaning x and y can be any real number.

The square root function y=√x rapidly increases at first but then flattens out. Its domain is [0, ∞) because x cannot be negative, and its range is also [0, ∞) because y values are never negative.

The natural logarithm function y=ln x is the inverse of e^x, having a vertical asymptote at x=0. Its domain is (0, ∞) and its range is (-∞, ∞). Inverse functions reflect across the line y=x.

The sine function y=sin x is a periodic wave that starts at the origin and oscillates between -1 and 1. Its domain is (-∞, ∞), and its range is [-1, 1].

The cosine function y=cos x is also a periodic wave, similar to sine but starting at its maximum value (1) when x=0. Its domain is (-∞, ∞), and its range is [-1, 1].

The tangent function y=tan x has vertical asymptotes at odd multiples of π/2. It is an increasing function. Its range is (-∞, ∞), but its domain excludes these asymptotes.

The video explains transformations. Multiplying a function f(x) by a constant outside the function (e.g., 2f(x)) results in a vertical stretch or shrink. 2f(x) stretches vertically by a factor of 2, while (1/2)f(x) shrinks vertically by a factor of 2.

Multiplying the x inside the function (e.g., f(2x)) results in a horizontal shrink or stretch. f(2x) causes a horizontal shrink by a factor of 2, and f(x/2) (f(0.5x)) causes a horizontal stretch by a factor of 2.

Adding or subtracting a constant inside the function (e.g., f(x-4)) causes horizontal shifts. f(x-4) shifts the graph 4 units to the right, and f(x+3) shifts it 3 units to the left.

A negative sign outside the function (-f(x)) reflects the graph over the x-axis. A negative sign inside the function (f(-x)) reflects the graph over the y-axis. -f(-x) reflects over the origin.

To graph an inverse function, simply swap the x and y coordinates of points on the original function. Inverse functions are reflections of each other across the line y=x.

For y=x²-3, the graph is a parabola shifted 3 units down. Domain remains (-∞, ∞), but the range is now [-3, ∞).

For y=-x²+2, the parabola opens downward and is shifted 2 units up. Domain is (-∞, ∞), and the range is (-∞, 2].

For y=(x-2)³, the cubic graph shifts 2 units to the right. Both domain and range remain (-∞, ∞).

For y=1/(x-3), the vertical asymptote shifts to x=3. The horizontal asymptote remains at y=0. Domain: (-∞, 3) U (3, ∞). Range: (-∞, 0) U (0, ∞).

For y=1/x+2, the horizontal asymptote shifts to y=2. The vertical asymptote remains at x=0. Domain: (-∞, 0) U (0, ∞). Range: (-∞, 2) U (2, ∞).

For y=-1/(x+2)+3, the vertical asymptote is at x=-2, the horizontal at y=3, and the negative sign reflects it. Domain: (-∞, -2) U (-2, ∞). Range: (-∞, 3) U (3, ∞).

For y=1/(x-2)²+3, the graph shifts 2 units right and 3 units up. It is symmetric about the vertical asymptote x=2. Domain: (-∞, 2) U (2, ∞). Range: (3, ∞).

For y=-1/(x+3)²-2, the vertical asymptote is x=-3, horizontal is y=-2, and it reflects vertically. Domain: (-∞, -3) U (-3, ∞). Range: (-∞, -2).

For y=|x-3|+1, the 'V' shape shifts 3 units right and 1 unit up. Domain: (-∞, ∞). Range: [1, ∞).

For y=-|x-2|+2, the 'V' shape shifts 2 units right and 2 units up, but opens downward. Domain: (-∞, ∞). Range: (-∞, 2].

For y=e^x+2, the horizontal asymptote shifts to y=2. Domain: (-∞, ∞). Range: (2, ∞).

For y=ln(x-3), the vertical asymptote shifts to x=3. Domain: (3, ∞). Range: (-∞, ∞).

For y=2sin x+1, the midline shifts to y=1, and the amplitude becomes 2. The graph oscillates between -1 and 3. Domain: (-∞, ∞). Range: [-1, 3].

For y=-3cos x+4, the midline shifts to y=4, amplitude is 3, and it reflects vertically. The graph oscillates between 1 and 7. Domain: (-∞, ∞). Range: [1, 7].

For y=³√x+1, the graph shifts 1 unit up. Both domain and range remain (-∞, ∞).

The video reviews different square root function orientations based on reflections (y=√x in Q1, y=-√x in Q4, y=√(-x) in Q2, y=-√(-x) in Q3).

For y=-√(x+2)+3, the origin shifts to (-2, 3), and the graph extends towards Q4 (right and down). Domain: [-2, ∞). Range: (-∞, 3].

For y=-√(-x+5)+3, the origin shifts to (5, 3), and the graph extends towards Q3 (left and down). Domain: (-∞, 5]. Range: (-∞, 3].

The video explains composite functions. For f(x)=x²+3 and g(x)=2x-4, f(g(x)) is found by substituting g(x) into f(x), resulting in (2x-4)²+3, which simplifies to 4x²-16x+19.

For the same functions, g(f(x)) is found by substituting f(x) into g(x), resulting in 2(x²+3)-4, which simplifies to 2x²+2.

If f(x)=3x-5 and g(x)=x³-9, to find f(g(2)), first calculate g(2) = 2³-9 = -1. Then find f(-1) = 3(-1)-5 = -8.

To find g(f(3)), first calculate f(3) = 3(3)-5 = 4. Then find g(4) = 4³-9 = 55.

To find the inverse of f(x)=7x-3, replace f(x) with y, swap x and y, then solve for y. This yields y = (x+3)/7.

Two functions f(x) and g(x) are inverses if f(g(x)) = x and g(f(x)) = x. The video demonstrates this proof for the previous example.

To find the inverse of f(x)=x² (for x≥0), replace f(x) with y, swap x and y, and solve for y. This results in y=√x. The graphs reflect across y=x.

The absolute value function y=|x| creates a 'V' shape opening upward. Its domain is (-∞, ∞), but its range is [0, ∞) as y values are always non-negative.

The rational function y=1/x has a horizontal asymptote at y=0 and a vertical asymptote at x=0. Its domain is (-∞, 0) U (0, ∞) (x ≠ 0), and its range is (-∞, 0) U (0, ∞) (y ≠ 0).

The function y=1/x² is similar to y=1/x but is symmetric about the y-axis, with both arms opening upward. The vertical asymptote is at x=0, and the horizontal asymptote is at y=0. Its domain is (-∞, 0) U (0, ∞), and its range is (0, ∞).

The exponential function y=e^x has a horizontal asymptote at y=0. It increases exponentially. Its domain is (-∞, ∞), and its range is (0, ∞).