Graphing of Functions Help

Example 1

Sketch the graph of f(x) = x ² .

Solution 1

Of course this is a simple and familiar function, and you know that its graph is a parabola. But it is satisfying to see calculus confirm the shape of the graph. Let us see how this works.

First observe that f′(x) = 2 x . We see that f′ < 0 when x < 0 and f′ > 0 when x > 0. So the graph is decreasing on the negative real axis and the graph is increasing on the positive real axis.

Next observe that f″ ( x ) = 2. Thus f″ > 0 at all points. Thus the graph is concave up everywhere.

Finally note that the graph passes through the origin.

We summarize this information in the graph shown in Fig. 3.4 .

Fig. 3.4

Example 2

Sketch the graph of f(x) = x ³ .

Solution 2

First observe that f′(x) = 3 x ². Thus f′ ≥ 0 everywhere. The function is always increasing.

Second observe that f″(x) = 6 x . Thus f″(x) < 0 when x < 0 and f″(x) > 0 when x > 0. So the graph is concave down on the negative real axis and concave up on the positive real axis.

Finally note that the graph passes through the origin.

We summarize our findings in the graph shown in Fig. 3.5 .

Fig. 3.5

Example 3

Sketch the graph of g(x) = x + sin x .

Solution 3

We see that g′(x) = 1 + cos x . Since −1 ≤ cos x ≤ 1, it follows that g′(x) ≥ 0. Hence the graph of g is always increasing.

Now g″(x) = − sin x . This function is positive sometimes and negative sometimes. In fact

− sin x is positive when kπ < x < ( k + 1) , k odd

and

− sin x is negative when kπ < x < ( k + 1) , k even.

So the graph alternates being concave down and concave up. Of course it also passes through the origin. We amalgamate all our information in the graph shown in Fig. 3.6.

Fig. 3.6

Example 4

Sketch the graph of h(x) = x /( x + 1).

Solution 4

First note that the function is undefined at x = − 1.

We calculate that h′ ( x ) = 1/(( x + 1) ² ). Thus the graph is everywhere increasing (except at x = −1).

We also calculate that h″ ( x ) = −2/(( x + 1) ³ ). Hence h″ > 0 and the graph is concave up when x < − 1. Likewise h″ < 0 and the graph is concave down when x > − 1.

Finally, as x tends to −1 from the left we notice that h tends to +∞ and as x tends to −1 from the right we see that h tends to −∞.

Putting all this information together, we obtain the graph shown in Fig. 3.7 .

Fig. 3.7

Sketch the graph of k(x) = x ³ + 3 x ² − 9 x + 6.

Solution 5

We notice that k′ ( x ) = 3 x ² + 6 x − 9 = 3( x − 1)( x + 3). So the sign of k′ changes at x = 1 and x = −3. We conclude that

k′ is positive when x < −3;

k′ is negative when −3 < x < 1;

k′ is positive when x > 3.

Finally, k″ ( x ) = 6 x + 6. Thus the graph is concave down when x < −1 and the graph is concave up when x > −1.

Putting all this information together, and using the facts that k(x) → −∞ when x → −∞ and k(x) → +∞ when x → +∞, we obtain the graph shown in Fig. 3.8 .

Fig. 3.8