Calculus I

Lesson 16: Analysing the Graphs of Functions 2 -
Extrema and Asymptotes

For each function, we find:

- intervals of increase or decrease

- extrema

- intervals of concavity

- points of inflection

- asymptotes

Example 1
f(x) = sin(2x) - 2 sin(x), x in [ - , ]

> restart: with(plots):

Warning, the name changecoords has been redefined

> f1:= x -> sin(2 * x) - 2 * sin(x);

> plot(f1(x), x=-Pi- .5..Pi + .5, color = red);

> D(f1);

We have: f1' (x) =

=

=

Hence we have critical values when cos (x) = OR cos (x) = 1.

Recall plot of cos (x).

> plot(cos(x), x= -Pi..Pi, color = red);

> solve(cos(x) = -1/2, x);

We see that our critical values are: x = 0, , and . Let's look at the plot of f1' (x).

Thus f1 is increasing on ( - , ) and ( , ).

Hence f1 has a local max when x = and a local min when x =

We now turn to concavity.

> D(D(f1));

Thus, f1'' (x) = -4 sin(2x) + 2sin(x)

= -4(2sin(x) cos(x))+ 2sin(x)

= 2sin(x) (-4cos(x) + 1)

Hence, f1'' (x) = 0 when sin(x) = 0 OR cos(x) = .

Since does not come from our standard triangles, we can only get a numerical estimate

for the solution to cos(x) = . Lets plot f '' (x).

> fsolve(cos(x) = 1/4, x = -Pi..Pi);

Conclusion, f 1 is concave up on ( ,0) and ( , )

and f1 is concave down on ( - , ) and (0, ).

There are points of inflection when x = - , - , 0, , .

There are no asymptotes.

Example 2

> f2:= x -> surd(x,3)^5 - 5 * surd(x,3)^2;

> plot(f2(x), x = -10..10, color = red);

> D(f2);

Too complcated!! We can do better ourselves!

f2 '(x) =

= .

Thus, f2 '(x) = 0 when x = 2 and f2 '(x) is not defined for x = 0.

Hence we have two critical points: x = 0, 2. Let's plot f2 '(x). Since f2 '(x) is not defined

at 0, we plot first with negative values of x and then with positive values of x.

Conclusion: f2 is increasing on (- ,0) and (2, ).

f2 is decreasing on (0,2).

Hence, f2 has a local max when x = 0 and a local min when x = 2.

The function is NOT differentiable at x = 0.

Now we turn to concavity.

> D(D(f2));

Again, too complicated! We can do better ourselves!

f2 ''(x) =

= .

We see that f2 ''(x) = 0 when x = -1 and is NOT defined when x = 0.

Let's plot f2 ''(x). Again we do two plots to avoid x = 0.

Conclusions: f2 is concave up on (-1,0) and (0, )

and f2 is concave down on (- ,-1).

There is a point of inflcetion when x = -1.

There are no asymptotes.

Example 3

> f3:= x -> x^2 / (2 * x + 5);

> plot(f3(x), x = -5..5,y = -25..25, color = red);

Using long division we have that

f3 = and hence

y = is a slant asymptote.

> D(f3);

> simplify( 2*x/(2 * x + 5) - 2 * x^2 / (2*x + 5)^2);

Hence there are critical points when x = 0, -5.

Note that is not a critical value as the function is not defined there.

To determine where f3 '(x) is positive and negative it suffices to consider ,

since is positive or 0.

> plot(x*(x+5), x = -10..10, color = red);

Hence, f3 is increasing on ( - ,-5) and (0, ).

The function f3 is decreasing on ( ) and ( ,0). Remember the

function is NOT defined for x = .

Thus, f3 has a local max when x = -5 and a local min when x = 0.

Let's turn to concavity.

> D(D(f3));

> simplify( 2*1/(2*x+5)-8*x/((2*x+5)^2)+8*x^2/((2*x+5)^3) );

Thus, f3 ''(x) = .

Conclusions: f3 is concave up on ( , ) and is concave down on (- , ).

Concavity chages as x passes through , but as is NOT in the domain

of the function we have no points of inflection.

> a3:= plot(f3(x), x = -10..10, y = -25..25,color = red):

> b3:= plot(1/2 * x - 5/4, x = -10..10, color = magenta):

> display({a3,b3});