This lesson is about deriving graphs of composite functions g(f(x)) from graphs of the inner function f(x).
The simplest case is finding the sum of two functions, f(x) and g(x). You can use the standard curve sketching process and also add ordinates of f and g at special points.
What constitutes a special point will depend on the situation. The process is:
Example To sketch of the graph of \[ y = x^2 + e^{-x} \quad \] First, sketch x2 and e-x
Intercepts: when x = 0, x2 + e-x = 1.
Asymptotes: when x is large and positive, x2 + e-x becomes large and positive. When x is large and negative, x2 + e-x becomes large and positive.
Special points: when x = 0, x2 + e-x = 1. When x = 1, x2 + e-x = 1 + 1/e. When x = -1, x2 + e-x = 1 + e
The above provides enough information for a sketch:
You can use the standard curve sketching process and also multiply ordinates of f and g at special points.
What constitutes a special point will depend on the situation. The process is:
Example To sketch of the graph of \[ y = x^2e^{-x} \ \] First, sketch x2 and e-x
Intercepts: when x = 0, x2e-x = 0
Asymptotes: when x is large and positive, x2e-x approaches 0. When x is large and negative, x2e-x becomes large and positive.
Special points: when x = 0, x2e-x = 0. When x = 1, x2e-x = 1/e. When x = -1, x2e-x = e. When x = 2, x2e-x = 0.54
The above provides enough information for a sketch:
Sketch \(y = \sin x e^{-x/(2\pi)}\) for x > 0.
The basic property of the absolute value function is that it reflects negative ordinates in the x-axis. Using this leads to the recipe:
Example Here is a sketch of the graph of \[ y = 1 + |x^2-3| \ \]
First, the parts of the parabola below the x-axis are reflected in the x-axis. Then the result is translated 1 unit in the positive y-direction.
Sketch \(y = |\sin x| - 1\).
In this situation, the absolute value on the argument of f converts f(-x) to f(x). This means the graph of x for x less than zero is the reflection of f for x > 0 in the y-axis. So:
Example Here is is a sketch of the graph of \[ y = \sin |x| \ \]
The green section is the reflection in the y-axis of the part of the graph for positive x.
Sketch \(y = \tan |x|\).
Guided Examples
O E Qs
Given f, we want to find \[ y = \frac{1}{f(x)} \ \]
The basic strategy here is to use the features of f to derive corresponding features of the reciprocal:
Example To sketch of the graph of \[ y = \frac{1}{\log x} \ \]
First, sketch log x
Intercepts: when x = 0, log x is not defined, so 1/log x does not intercept the y-axis. When y = 0, log x intercepts the x-axis at x = 1. But 1/log 1 is not defined, so 1/log x does not intercept the x-axis
Asymptotes: when x is large and positive, 1/log x approaches zero. From the intercept analysis, x = 1 is a vertical asymptote.
Special points: log x = 1/log x when x = e and 1/e. So the intersection points are (e, 1) and (1/e, -1)
As x approaches zero from the right, log x becomes unbounded, so 1/log x is not defined at x = 0.
The above provides enough information for a sketch:
Sketch \(y = \dfrac{1}{x^2-2x}\).
In this situation, you can use the properties of the square root function.
Example Here is a sketch of \[ y = \sqrt{2\sin x} \quad \ \]
Sketch \(y = \sqrt{x^3 - x}\).
This means you want to find the function y such that y2 = f(x).
Consider the simplest case \( y^2 = x \). This is a parabola with the x-axis as axis of symmetry and vertex at y = 0. To sketch the curve you can set y = ± \( \sqrt x \) and then sketch both square roots. However, the square root function is symmetric about the x-axis so you just need to sketch the positive square root and reflect the result in the x-axis to get the rest of the curve.
So the process is:
Example To sketch \[ y^2 = x^3 - 3x^2 + 2x = x(x-1)(x-2) \ \]
First, sketch f(x)
So \[ y = \pm \sqrt{x(x-1)(x-2)} \ \] For the positive part of y,
Intercepts: y = 0 when x = 0 or 1 or 2.
Asymptotes: when x is large and positive, y is large and positive. y is not defined for x less than 0.
Vertical tangents occur when y = 0. This is when x = 0, 1 or 2.
The above provides enough information for a sketch:
The reflection about the x-axis is in green.
Sketch \(y^2 = x^2(x-1)\).