2 B The Exponential Function e x and the Definition of e

This section introduces an important new number callede. This number e is not a whole number, or even a fraction, but is a real number between 2 and 3 whose value is approximately e =.. 2·7183.

Differentiating exponential functions whose base is a whole number, likey = 2^x or y = 10^x, turns out to be inconvenient. It is much easier to use this new numbere as a base and to study the exponential function y = e^x. The fundamental result of this section is that the functiony = e^x is its own derivative:

d

dxe^x=e^x. Differentiatingy = 2^x:

We begin by looking aty = 2^x and trying to differentiate it. Below is a sketch of y = 2^x, with the tangent drawn at itsy-intercept A(0, 1).

Differentiating 2^x requires first-principles differentiation, because the theory so far hasn’t provided any rule for differentiating 2^x.

The formula for first-principles differentiation is f^(x) = lim

h→0

f(x + h) − f(x)

h .

Applying this formula to the functionf(x) = 2^x, f^(x) = lim

h→0

2^x+h − 2^x h

x y

1 1 2

= lim A

h→0

2^x× 2^h− 2^x

h , since 2^x× 2^h = 2^x+h, and taking out the common factor 2^x in the numerator,

f^(x) = 2^x× lim

h→0

2^h− 1 h

= 2^x× m, where m = lim

h→0

2^h− 1 h .

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This limitm = lim

h→0

2^h− 1

h cannot be found by algebraic methods,

but substitutingx = 0 gives a very simple geometric interpretation of the limit:

f^(0) = 2⁰× m

=m, since 2⁰ = 1, som = lim

h→0

2^h− 1

h is just the gradient of the tangent toy = 2^x at its y-intercept (0, 1).

The conclusion of all this is that d

dx2^x= 2^x× m, where m is the gradient of y = 2^x at its y-intercept.

Exactly the same argument can be applied to any exponential function y = a^x simply by replacing 2 bya in the calculation above:

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DIFFERENTIATINGy = a^x: For all positive numbers a, d

dxa^x=a^x× m, where m is the gradient of y = a^x at its y-intercept.

That is, the derivative of an exponential function is a multiple of itself.

The Definition ofe: It now makes sense to choose the base that will make the gradient exactly 1 at they-intercept, because the value of m will then be exactly 1. This base is given the symbole and has the value e =.. 2·7183.

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THE DEFINITION OFe: Define e to be the number such that the exponential function y = e^x has gradient exactly 1 at itsy-intercept. Then

e =.. 2·7183.

The function y = e^x is called the exponential function to distinguish it from all other exponential functionsy = a^x.

A calculator will provide approximate values ofe^xcorrect to about ten significant figures — use the function labelled e^x , which is usually located above the button labelled ln .

Sincee = e¹, an approximation for the numbere itself is found using the function labelled e^x , using the input x = 1.

WORKEDEXERCISE:

Use your calculator to find, correct to four significant figures:

(a) e² (b) e (c) 1

e (d) √

e SOLUTION: Using the function labelled e^x on the calculator:

(a) e² =.. 7·389 (b) e = e¹

=.. 2·718

(c) 1 e =e⁻¹

=.. 0·3679

(d) √ e = e¹²

=.. 1·649

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The Derivative ofe^x: The fundamental result of this section then follows immediately from the previous two boxed results.

d

dxe^x =e^x× m, where m is the gradient of y = e^x at its y-intercept,

=e^x× 1, by the definition of e,

=e^x.

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THE DERIVATIVE OF THE EXPONENTIAL FUNCTIONy = e^x: The exponential functiony = e^x is its own derivative:

d

dxe^x =e^x.

The graph of y = e^x has been drawn below on graph paper. The tangent has been drawn at they-intercept (0, 1) — this shows that the gradient of the curve there is exactly 1.

This graph of y = e^x is one of the most important graphs in the whole course and its shape and properties need to be memorised.

x y

0 2 3

−2 1

1e

e

−1

1

y x= + 1

y e= ^x

• The domain is all values of x.

The range is y > 0.

• There are no zeroes.

The curve is always above the x-axis.

• As x → −∞, y → 0.

This means that thex-axis y = 0 is a horizontal asymptote to the curve on the left-hand side.

• As x → ∞, y → ∞.

On the right-hand side, the curve rises steeply.

• The curve has gradient 1 at its y-intercept (0, 1).

• The curve is always concave up.

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Gradient Equals Height: The fact that the derivative of the exponential functione^x is the same function has a striking geometrical interpretation in terms of its graph.

Ify = e^x, then dy

dx =e^x, which means that for this function dy

dx =y.

Thus at each point on the curvey = e^x, the gradient dy

dx of the curve is equal to the heighty of the curve above the x-axis. We have already seen this happening at they-intercept (0, 1), where the gradient is 1 and the height is also 1.

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GRADIENT EQUALS HEIGHT:

At each point on the graph of the exponential functiony = e^x, dy

dx =y.

That is, the gradient of the curve is always equal to its height above the x-axis.

This property of the exponential function is the reason why the function is so important in calculus. An important question in the next exercise asks for this to be verified on a graph-paper graph ofy = e^x.

Transformations of the Exponential Graph: The usual methods of shifting and reflect-ing graphs can be applied to y = e^x. When the graph is shifted vertically, the horizontal asymptote aty = 0 will be shifted also.

A small table of approximate values can be a very useful check, particularly when a sequence of transformations is involved.

WORKEDEXERCISE:

Use transformations of the graph ofy = e^x, and a table of values, to generate a sketch of each function. Show they-intercept and the horizontal asymptote and state the range.

(a) y = e^x+ 3 (b) y = e^−x (c) y = e^x−2 SOLUTION:

(a) Graphy = e^x+ 3 by shiftingy = e^x up 3 units.

x −1 0 1

y e⁻¹+ 3 4 e + 3

approximation 3·37 4 5·72

y-intercept: (0, 4) Asymptote: y = 3

x y

1 3 4 e + 3

Range: y > 3

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(b) Graphy = e^−x by reflecting y = e^x in the y-axis.

x −1 0 1

y e 1 e⁻¹

approximation 2·72 1 0·37 y-intercept: (0, 1)

Asymptote: y = 0 (the x-axis)

x y

−1 1 e

Range: y > 0

(c) Graphy = e^x−2 by shiftingy = e^x to the right by 2 units.

x 0 1 2 3

y e⁻² e⁻¹ 1 e

approximation 0·14 0·37 1 2·72 y-intercept: (0, e⁻²)

Asymptote: y = 0 (the x-axis) ^x

y

2 1

e⁻²

Range: y > 0

Bounds fore: It’s not appropriate in this course to spend time calculating close ap-proximations toe. The graphs below at least show very quickly that the number e lies between 2 and 4.

Here are tables of values and sketches ofy = 2^x and y = 4^x. On each graph, the tangent at the y-intercept A(0, 1) has been drawn.

x y

1 1

2

A

B

12

12

− ¹2 x

y

A C

1 2

For y = 2^x, x 0 1

y 1 2

LetB = (1, 2).

Then the chordAB has gradient 1, so the tangent at A must

have gradient less than 1.

Fory = 4^x, x −¹₂ 0

y ¹₂ 1

LetC = (−¹₂,¹₂).

Then the chordCA has gradient 1, so the tangent at A must

have gradient greater than 1.

These two diagrams show that the tangent at they-intercept A(0, 1) has gradient less than 1 fory = 2^x and greater than 1 fory = 4^x. Since the gradient ofy = e^x at its y-intercept is exactly 1, it follows that

2< e < 4.

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Exercise 2B

1. Use the function labelled e^x on your calculator to approximate the following correct to four decimal places:

(a) e² (b) e³ (c) e¹⁰

(d) e⁰ (e) e¹ (f) e⁻¹

(g) e⁻² (h) e¹²

(i) e⁻¹²

x y

0 1 2 3

−1

−2 1

e1

e 2.

These questions refer to the graph ofy = e^x drawn above .

(a) Photocopy the graph ofy = e^x above and on it draw the tangent at the point (0, 1) where the height is 1, extending the tangent down to thex-axis.

(b) Measure the gradient of this tangent and confirm that it is equal to the height of the exponential graph at the point of contact.

(c) Copy and complete the table of values below by measuring the gradient y^ of the tangent at the points where the heighty is ¹₂, 1, 2 and 3.

height y ¹₂ 1 2 3

gradienty^ gradient

height

(d) What do you notice about the ratios of gradient to height?

3. (a) Photocopy the graph of y = e^x in question 2 and on it draw the tangent at (0, 1), extending the tangent down to thex-axis.

(b) Measure the gradient of this tangent and confirm that it is equal to the height of the graph at (0, 1).

(c) Draw the tangents at the three points wherex = −2, x = −1 and x = 1. Measure the gradient of each of these tangents and confirm that it is equal to the height of each point of contact.

(d) What do you notice about thex-intercepts of all the tangents?

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−1 0 1

1 2 3

x 4. y

(a) Photocopy the graph ofy = 2^x and on it draw tangents atx = −1, 0, 1 and 2.

(b) Copy and complete the table of values to the right by measuring the gradient y^ of each tangent.

(c) What do you notice about the ratios of gradient to height?

x −2 −1 0 1 2

height y gradient y^

gradient height

D E V E L O P M E N T

5. Sketch the graph of y = e^x, then use your knowledge of transformations to graph the following functions. Show the horizontal asymptote and state the range in each case.

(a) y = e^x+ 1 (b) y = e^x+ 2 (c) y = e^x− 1 (d) y = e^x− 2 6. Sketch the graph of y = e^−x, then use your knowledge of transformations to graph the

following functions. Show the horizontal asymptote and state the range in each case.

(a) y = e^−x+ 1 (b) y = e^−x+ 2 (c) y = e^−x− 1 (d) y = e^−x− 2 7. (a) What is the y-coordinate of the point on the curve y = e^x wherex = 0?

(b) Use the result d

dxe^x =e^x to find the gradient of the tangent at this point.

(c) Hence write down the equation of the tangent, and find itsx-intercept.

(d) Repeat the above steps for the points where x = −2, −1 and 1.

(e) Compare the values of the x-intercepts with those found in question 3.

8. (a) Sketch a graph ofy = e^x and hence write down its range.

(b) Write down y^ and hence explain why the graph always has positive gradient.

(c) Write down y^ and hence explain why the graph is always concave up.

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9. (a) Copy and complete the following tables of values for the functionsy = e^x andy = e^−x, giving your answers correct to two decimal places.

x −2 −1 0 1 2

e^x

x −2 −1 0 1 2

e^−x

(b) Sketch both graphs on one number plane, and draw the tangents at eachy-intercept.

(c) Find the gradients of the two tangents, and hence explain why they are perpendicular.

10. Use the graph ofy = e^x and your knowledge of transformations to graph:

(a) y = e^x−1 (b) y = e^x−3

(c) y = e^x+1 (d) y = e^x+2

(e) y = −e^x (f) y = e^−x

11. (a) Photocopy the graph ofy = 2^x in question 4 and on it draw the chord fromx = −1 tox = 0, and another from x = 0 to x = 1. Find the gradients of these chords.

(b) Draw the tangent at x = 0 and compare it with the chords. Hence explain why the gradient of the tangent lies between ¹₂ and 1.

(c) Measure the gradient of the tangent to confirm this.

C H A L L E N G E

12. [Technology]

(a) A graphing program can be used to draw tangents to a graph of y = e^x at various points on the curve and confirm that at each point the gradient equals the height.

This exercise was done on graph paper in questions 2 and 3 above.

(b) The transformations in questions 5, 6, 10, 16 and 17 of the graph of the exponential functiony = e^x can be confirmed using a graphing program, after which experimen-tation with further transformed graphs can be carried out.

13. [Technology] On page 65 it was shown from first principles that the gradient of y = 2^x at its y-intercept is given by lim

h→0

2^h − 1

h . Use a calculator or computer to evaluate 2^h− 1 h , correct to five decimal places, for the following values ofh:

(a) 1 (b) 0·1 (c) 0·01 (d) 0·001 (e) 0·0001 (f) 0·000 01 14. [Technology]

(a) Use a graphing program to graph y = 2^x and y = x + 1 on the same number plane, and hence observe that the gradient ofy = 2^x at (0, 1) is less than 1.

(b) Similarly, graphy = 3^x and y = x + 1 on the same number plane and hence observe that the gradient ofy = 3^x at (0, 1) is greater than 1.

(c) Choose a sequence of bases between 2 and 3 so that the gradient at (0, 1) converges to exactly 1. In this way a reasonable approximation fore can be obtained.

15. Use the graph of y = e^x and your knowledge of transformations to graph the following functions. Show the horizontal asymptote and state the range in each case.

(a) y = 1 − e^x (b) y = 3 − e^x (c) y = − e^−x (d) y = − e^x−1 16. The function e^x may be approximated using the power series

e^x = 1 + x

1 + x²

1× 2 + x³

1× 2 × 3 + x⁴

1× 2 × 3 × 4 + · · · .

Use this power series to approximate each of the following, correct to two decimal places.

Then compare your answers with those given by your calculator.

(a) e (b) e⁻¹ (c) e² (d) e⁻²

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