Análisis de la ejecución del gasto en proyectos por Jurisdicción

4.4. Curve sketching

Classifying stationary points: We can also determine whether each of the stationary points is a local maximum, local minimum or point of inflection by using the

methods outlined above.

Limiting behaviour in the x-direction: We can determine how f (x) is behaving as x→ ∞ and as x → −∞.

Of course, in certain cases, it may also be advantageous to think carefully about the intervals in which the function is increasing (or decreasing) or whether the function is convex (or concave). But, generally, the method above should suffice when we sketch most functions.

In particular, observe that a sketch is very different from a plot. A plot involves plotting certain points and joining them up with little regard to any interesting behaviour the curve may be exhibiting elsewhere. A sketch, on the other hand, isolates any interesting behaviour the curve may be exhibiting (such as the ones listed above) and concentrates on these. Please be aware that there is a difference and in this course, we will always want to see sketches and not plots!

To see how we can implement the method above, we will start by sketching the relatively simple curves that arise when f is a polynomial. We will then consider how we would proceed when the functions are differentiable, but involve other elementary functions. Then, just so that we are aware of some possible complications, we look at what happens when our function fails to be differentiable at some points.

4.4.1 Sketching curves defined by polynomials

Given what we have seen so far, the only real obstacle to sketching a polynomial is an understanding of the limiting behaviour of this kind of function. The key result here is that, if f (x) is a polynomial, its behaviour as x gets arbitrarily large in magnitude (that is to say, as x→ ∞ or x → −∞) is determined solely by its leading term, i.e. the one with the highest power of x. Then, with this in mind, we can look at the term with the highest power of x, let’s say that this is xⁿ, and note that:

if n is even, then xⁿ→ ∞ as x → ∞ and as x → −∞; whereas if n is odd, then xⁿ→ ∞ as x → ∞ and xⁿ → −∞ as x → −∞.

Using these facts and noting how the sign of the coefficient of the term with the highest power of x can influence the sign of the limit, we can determine the limiting behaviour of any polynomial.

Activity 4.3 Suppose that f (x) is a polynomial and that, for some constants a6= 0 and n∈ N, the term in this polynomial with the highest power of x is axⁿ.

Determine the behaviour of f (x) as x→ ∞ and as x → −∞ in the cases which arise according to whether a is positive and negative and whether n is even or odd.

We can now see how to sketch some polynomials and we start by seeing how to sketch the function that we have been considering throughout this chapter.

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Example 4.7 Sketch the curve y = f (x) where f (x) is the function in Example 4.1.

From the earlier examples in this section, we know quite a lot about this function and, in particular, we have found and classified its stationary points. But, to sketch this curve, we need to find a bit more information, namely its

x-intercepts: These occur when y = 0 and so we solve the equation given by f (x) = 0, i.e.

x³− 2x²− 15x = 0,

which, on taking out the common factor of x and factorising the remaining quadratic, gives us

x(x²− 2x − 15) = 0 =⇒ x(x− 5)(x + 3) = 0.

Thus, the x-intercepts occur when x =−3, x = 0 and x = 5.

y-intercept: This occurs when x = 0 and so using y = f (0) we see that the y-intercept occurs when y = 0. Note, in particular, that this means that the curve goes through the origin (as we should have expected since one of the x-intercepts occurs when x = 0).

stationary points: We have found the x-coordinates of the stationary points and classified them above (see, for instance, Example 4.2). So, all we need to do here, is use y = f (x) to find the values of y at these points so that we can locate them on our sketch. Doing this, we find that f (x) has a

• local maximum when x = −5/3 and y = f(−5/3) = 400/27, and

• local minimum when x = 3 and y = f(3) = −36.

limiting behaviour: The term with the highest power of x in f (x) is x³ and so f (x) → ∞ as x → ∞ and f(x) → −∞ as x → −∞.

So, using this information, we begin to sketch this curve by roughly indicating these key features on some axes as in Figure 4.9(a) and then, joining them up with a nice smooth curve, we get the sketch itself as in Figure 4.9(b).

In particular, it is worth noting that in this sketch:

all of the key features are labelled;

the curve has the right kind of limiting behaviour, i.e. f (x)→ ∞ as x → ∞ and f (x) → −∞ as x → −∞; and

points of inflection which are not stationary points (recall that, in Example 4.5, we saw that this curve has one when x = 2/3) are not usually indicated.

Of course, what we see here is similar to what we saw in Figure 4.2, but a sketch must include information about all of the relevant key features.

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4.4. Curve sketching

O x

y

−⁵₃

40027

−36

−3

3

5 O x

y

−⁵₃

y = f (x)

400 27

−36

−3

3

5

(a) The key features (b) The sketch

Figure 4.9: Sketching the curve y = x³− 2x² − 15x in Example 4.7. (a) Using what we have discovered about the key features of the curve, we can begin to see what it must look like. (b) By joining up these key features with a nice smooth curve, we get the sketch itself.

Indeed, it can be seen that, unlike plotting a function, sketching it is a bit of an art and it can only be done well by learning to appreciate what your calculations are telling you about its appearance. With this in mind, let’s sketch a function that we haven’t

encountered before.

Example 4.8 Sketch the curve y = f (x) where f (x) = 2x⁴− 4x³+ 2x². We find the key features of this curve according to the list given above, namely

x-intercepts: These occur when y = 0 and so we solve the equation given by f (x) = 0, i.e.

2x⁴− 4x³+ 2x² = 0,

which, on taking out the common factor of 2x² and factorising the remaining quadratic, gives us

2x²(x²− 2x + 1) = 0 =⇒ 2x²(x− 1)² = 0.

Thus, the x-intercepts occur when x = 0 and x = 1.

y-intercept: This occurs when x = 0 and so using y = f (0) we see that the y-intercept occurs when y = 0. Note, in particular, that this means that the curve goes through the origin (as we should have expected since one of the x-intercepts occurs when x = 0).

finding the stationary points: These occur when f⁰(x) = 0 and so, noting that f⁰(x) = 8x³− 12x²+ 4x,

we solve the equation

8x³− 12x²+ 4x = 0,

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which, on taking out a common factor of 4x and factorising the remaining quadratic, gives us

4x(2x² − 3x + 1) = 0 =⇒ 4x(2x− 1)(x − 1) = 0,

and so the stationary points occur when x = 0, x = 1/2 and x = 1. Then, we use y = f (x) to find the values of y at these points so that we can locate them on the sketch. Doing this, we find that

• x = 0 gives y = f(0) = 0,

• x = 1/2 gives y = f(1/2) = 1/8, and

• x = 1 gives y = f(1) = 0.

So, the stationary points have coordinates given by (0, 0), (1/2, 1/8) and (1, 0).

classifying the stationary points: Let’s use the second-order derivative test here.

We can see that

f⁰⁰(x) = 24x²− 24x + 4, and so, looking at the stationary points, we have

• f⁰⁰(0) = 4 > 0 and so (0, 0) is a local minimum;

• f⁰⁰(1/2) =−2 < 0 and so (1/2, 1/8) is a local maximum; and

• f⁰⁰(1) = 4 > 0 and so (1, 0) is a local minimum.

limiting behaviour: The term with the highest power of x in f (x) is 2x⁴ and so f (x) → ∞ as x → ∞ and as x → −∞.

So, using this information, we begin to sketch this curve by roughly indicating these key features on some axes as in Figure 4.10(a) and then, joining them up with a nice smooth curve, we get the sketch itself as in Figure 4.10(b).

y

O ¹₂ 1 x

1 8

y

O 1 x

y = f (x)

1 2 1

8

(a) The key features (b) The sketch

Figure 4.10:Sketching the curve y = 2x⁴− 4x³+ 2x² in Example 4.8. (a) Using what we have discovered about the key features of the curve, we can begin to see what it must look like. (b) By joining up these key features with a nice smooth curve, we get the sketch itself.

Activity 4.4 Find the points of inflection of the function in Example 4.8.