Page 191 - DMTH201_Basic Mathematics-1

Basic Mathematics – I

Notes The range of f is the interval (-inf , +inf).
2. The vertical asymptote is obtained by solving
x + 2 = 0
which gives

x = -2
As x approaches -2 from the right (x > -2), f(x) decreases without bound. How do we know
this?

Let us take some values:
2
2
f(-1) = log (-1 + 2) = log (1) = 0
2
2
f(-1.5) = log (-1.5 + 2) = log (1/2) = -1
2
f(-1.99) = log (-1.99 + 2) = log (0.01) which is approximately equal to -6.64.
2
f(-1.999999) = log (-1.999999 + 2) = log (0.000001) which is approximately equal
2
2
to -19.93.
3. To find the x intercept we need to solve the equation f(x) = 0
log (x + 2) = 0
2
Use properties of logarithmic and exponential functions to write the above equation as:
2
2log (x + 2) = 20
Then simplify

x + 2 = 1
x = -1
The x intercept is at (-1, 0).
The y intercept is given by (0, f(0)) = (0, log (0 + 2)) = (0, 1).
2
4. So far we have the domain, range, x and y intercepts and the vertical asymptote. We need
more points. Let us consider a point at x = -3/2 (half way between the x intercept and the
vertical asymptote) and another point at x = 2.
2
2
f(-3/2) = log (-3/2 + 2) = log (1/2) = log (2 - 1) = -1.
2
2
f(2) = log (2 + 2) = log (22) = 2.
2
We now have more information on how to graph f. The graph increases as x increases.
Close to the vertical asymptote x = -2, the graph of f decreases without bound as x approaches
-2 from the right. The graph never cuts the vertical asymptote. We now join the different
points by a smooth curve.

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