Límites del Derecho a la Libertad -Libertinaje, Abuso de Ejercicio del

Solution

When x = 0, y = 0.e⁰ = 0 × 1 = 0, so the tangent passes through the point (0, 0).

The gradient of the tangent is the gradient of the curve, and dy

dx = e^x.1 + x.e^x = e^x(1 + x) so when x = 0

dy

dx = m = e⁰(1 + 0) = 1 so our line has the form

y = 1.x + c

and inserting the point we know is on the line (0, 0) gives 0 = 1.0 + c ⇒ c = 0

and thus our equation is

y = x

9.6 Turning Points

There are certain features in curves that particularly interest us. An example would be the “peaks” and “troughs” of a curve, or in more exact language, the “maxima” and “minima”. These, together with two other cases we shall see later, are known as turning points.

Looking at such points on a curve reveals an important fact: at just the point of the maximum or minimum, the curve is horizontal. In other words the gradient is zero at these points.

9.6.1 Types of turning point

There are four types of turning point, or points where the gradient is zero for an instant. These are shown in figure 9.1.

Local maximum

A local maximum is a point on the curve so that if we were to imagine standing at that point, the curve would descend in both directions, at least for a little distance. This is case 1 in figure 9.1.

9.6 Turning Points 159

Figure 9.1: Types of turning point

This local maximum may not be the same as the maximum point of the curve (there may be some other peak which is higher).

Local minimum

A local minimum is a point on the curve so that if we were to imagine standing at that point, the curve would ascend in both directions, at least for a little distance. This is case 2 in figure 9.1.

This local minimum may not be the same as the minimum point of the curve (there may be some other trough which is deeper).

Stationary points of inflection

A stationary point of inflection is a point on the curve so that if we were to imagine standing at that point, it would be flat at that exact point, the curve would ascend in one direction and descend in the other, at least for a little distance.

There are two varieties of this turning point, one that goes down to the left and up to the right (this is case 3 in figure 9.1), and one which goes up to the left and down to the right (this is case 4 in figure 9.1).

9.6.2 Finding turning points

To find the turning points of a curve y = f (x) we first of all find the derivative, f⁰(x) and solve the equation

9.6 Turning Points 160

f⁰(x) = dy dx = 0

for x. There may be no, one or several values of x that solve this equa-tion. For each value of x, place it into the formula for y = f (x) to find the corresponding y coordinates.

In doing this we have found all the turning points of the curve.

9.6.3 Classification of turning points

Once we have located our turning points we usually wish to know what type of turning point each one is. There are two ways of determining this.

Second derivative test

The first method is to differentiate the derivative again, to obtain the second derivative.

We plug the x values from each turning point into this function and, depending upon the sign of our answer we can determine which type of turning point we have. This is shown in table 9.1.

f⁰⁰(x) Turning Point Comments + Local minimum Conclusive

− Local maximum Conclusive 0 Stationary inflection Inconclusive

Table 9.1: Second derivative test

Note that if the second derivative is zero, we know that we have a sta-tionary point of inflection, but not which case. We shall need to use another test in this case.

Direct analysis of the gradient

This method is usually more difficult to employ, but in some cases it can be easier, such as when f⁰(x) is a complex fraction and differentiating again will be very arduous. Alternatively, we may be forced to use this method when the second derivative test is inconclusive.

9.6 Turning Points 161

To use this method we examine the sign of the gradient, f⁰(x) a little to the left and to the right of the turning point. It is important we stay close however. The results we can conclude are shown in table 9.2.

Left Right Conclusion

+ − Local maximum

− + Local minimum

+ + Stationary inflection, bottom left to top right

− − Stationary inflection, top left to bottom right

Table 9.2: First derivative turning point classification

9.6.4 Example

Find and classify the turning points of the function y = x³ + 3x²− 24x + 3.

Solution

Find the first and second derivatives dy

dx = 3x²+ 6x − 24;d²y

dx² = 6x + 6.

We find the turning points by solving dy

dx = 0 ⇒ 3x² + 6x − 24 = 0, and this is simply a quadratic equation

⇒ x²+ 2x − 8 = 0 ⇒ (x + 4)(x − 2) = 0 ⇒ x = −4, x = 2

so there are two turning points. We find the matching y coordinates for each x value

x = −4 ⇒ y = (−4)³+ 3(−4)²− 24(−4) + 3 = 83 and

x = 2 ⇒ y = 2³+ 3(2)²− 24(2) + 3 = −25.

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Finally it simply remains to classify the points, using the second derivative test.

x = −4 ⇒ d²y

dx² = 6(−4) + 6 = −ve which indicates a local maximum, and

x = 2 ⇒ d²y

dx² = 6(2) + 6 = +ve which is a local minimum.

Thus we have (−4, 83), a local maximum and (2, −25), a local minimum.

9.6.5 Example

Find and classify the turning points of the function y = x³.

Solution

Find the first and second derivatives dy

dx = 3x²;d²y dx² = 6x.

Now solve to find turning points dy

dx = 0 ⇒ 3x² = 0 ⇒ x = 0

so we have only one turning point at x = 0. Now we classify it, trying the second derivative test first.

x = 0 ⇒ d²y

dx² = 6(0) = 0 which is inconclusive.

We have to fall back on using the first derivative. Try just left of the turning point, say when x = −1

dy

dx = 3(−1)² = 3 = +ve and right just a bit, say when x = 1

dy

dx = 3(1)² = 3 = +ve