Evaluation Test - Functions
CHAPTER 3 SOLUTIONS TO REINFORCEMENT EXERCISES IN FUNCTIONS AND SERIES
3.3.1.Definition of a function
3.3.1A.
Find the values of the following functions at the points a) – 1, b) 0, c) 3 d) – .
i)f(x) = 2xii)f(x) = 3x2 – 1
iii)f(x) = 2x2 – x + 1iv)f(x) = x – 2
v)f(x) = x –
Solution
i)If f(x) = 2x then
a) f(–1) = 2(– 1) = –2
b) f(0) = 2 (0) = 0
c) f(3) = 2(3) = 6
d) f= 2 = –
ii)If f(x) = 3x2 – 1 then
a) f(– 1) = 3(– 1)2 – 1 = 3 – 1 = 2
b) f(0) = 3(0)2 – 1 = 0 – 1 = – 1
c) f(3) = 3(3)2 – 1 = 27 – 1 = 26
d) f= 32 – 1 = – 1 = –
iii)If f(x) = 2x2 – x + 1 then
a) f(– 1) = 2(– 1)2 – (– 1) +1 = 2 + 1+ 1= 4
b) f(0) = 2(0)2 – 0 + 1 = 1
c) f(3) = 2(3)2 – 3 + 1 = 18 – 3 + 1 = 16
d) f= 2 2 – + 1 = + + 1 = + + =
iv) If f(x) = then
a) f(– 1) = = 1
b) f(0) = =
c) f(3) = =
d) f= = =
v)If f(x) = then
a) f(– 1) = = = 4
b) f(0) = = = – 1
c) f(3) = =
d) f= = = – 6
3.3.1B.
If f(x) = obtain expressions for
i)f(u)ii)f(t + 1)iii)f
iv) f(a + h)v)f(x + x)
Solution
In each case we have to replace the x in f(x) with the corresponding given expression. If f(x) = then
i) f(u) =
ii)f(t + 1) = = =
iii)f= = /=
iv) f(a + h) = =
v)f(x + x) = =
3.3.2Plotting the graph of a function
3.3.2A.
Plot the graphs of the following functions over the ranges indicated
i)2x3 – 3 x 3 ii) – 6 x 6 iii) 2x – 4 x 4
Solution
In each case plotting values at, say, integer intervals (except for x = 0 in the case of ii) of course!) between the extremes given and drawing a continuous curve through the resulting points should give the curves shown below. Note that although we would normally like to use the same scales on each axis (as, for example, in Figures 3.1 and 3.2 of UEM) you will soon discover that for these functions this is not really practical. It is useful to think about why this is so in each case, In i) we have a strong power function and clearly it changes very rapidly as x changes. In ii), no matter what the value of x, the bulk of the change in the function is going to occur when x is near the origin, because the reciprocal means that as x gets smaller and smaller in magnitude, the function increases rapidly. At x = 0 the function 3/x does not exist of course and you may need to plot a few values less than 1 to convince you of what happens - the curve shoots off to '+ infinity' (x > 0) or ' infinity'. This is pretty clear from noting that if n is any integer (non-zero) then for x = 1/n, y = 3/x becomes y = 3n. Then as n gets larger, x gets smaller and y gets larger. iii) is an early introduction to an exponential function (see UEM 4). You can see that as x increases such a function increases even more rapidly than the power function in i) as x gets larger.
A. (i) (ii)
(iii)
3.3.2B.
Plot the graph of the function f(x) = 2x2 – 3x – 2 and confirm the statement made in Section 3.2.6 about the range of values of x for which this function is positive.
Solution
A few plotted points should reveal the form of this quadratic quite quickly. In this case, because we can solve the quadratic equation 2x2 – 3x – 2 = 0 explicitly, we can determine the precise points where the curve crosses the x - axis, namely at x = and x = 2. This defines precisely the limits outside of which the function is positive. If we could not solve the equation in this way then we would have to estimate where the curve crosses the x-axis, which would require a more accurate graph.
B.
3.3.3Formulae
The following are examples of standard formulae occurring in engineering, with standard notation. In each case define the variables, explain what the formula tells us, and describe the type of function that the left hand side is of the bracketed variables. Make the indicated variable the subject of the formula.
i)V = IR (R)ii)P = I2R (R)
iii)E = mv2 (v)iv)s = ut + at2 (t)
v)E = mc2 (c)
Solution
i) V = IR is Ohm's law relating voltage V to current I and resistance R in a wire. To put R in terms of V and I, divide through by I to get
R =
ii)P = I2R gives the power P of a resistance R passing current I. Solving for R gives
R =
iii) E = mv2 gives the kinetic energy E of a particle with mass m moving in one dimension with velocity v. Solving for v we have
v2 = so v =
iv) s = ut + at2 describes the distance s travelled in time t by a particle moving in one dimension with constant acceleration, starting from rest with initial velocity u. This is a quadratic in t and we can solve it using the formula:
t =
v)E = mc2 is said to be the most famous equation known, and is the expression for the (rest) energy equivalent to a rest mass m, where c is the velocity of light. There is at least one book all about this equation (E = Mc2by David Bodanis). Solving for c we have
c2 =
so c = (the positive value)
3.3.4Odd and even functions
State whether the following functions are even, odd or neither.
i)2xii)3x2 – 1
iii)2x3 – xiv)x2 + 2x + 1
v)cos xvi)x4 + 2x2 + 1
vii)sin xviii)ex
Solution
i) If f(x) = 2x then f(– x) = 2(– x) = – 2x = – f(x) so this function is odd
ii) With f(x) = 3x2 – 1 we have f(– x) = 3(– x)2 – 1 = 3x2 – 1 = f(x), so f(x) is even
iii)f(x) = 2x3 – x is odd, using the same arguments as in i), or simply note that it has only odd powers of x.
iv)If f(x) = x2 + 2x + 1 then f(– x) = (– x)2 + 2(– x) + 1 = x2 – 2x + 1. This is neither f(x) nor – f(x) and so the function is neither odd nor even
v)As we will see in Chapter 6 cos x is an even function: cos(– x) = cosx
vi)Either note that x4 + 2x2 + 1 only has even powers of x, or note explicitly that (– x)4 + 2(– x)2 + 1 = x4 + 2x2 + 1. Either way the function is even
vii)sin x is an odd function, sin(– x) = – sinx
viii)ex is neither odd nor even. Thus if f(x) = ex then f(– x) = e– x which is not equal to f(x) or – f(x) (See Chapter 4).
3.3.5Composition of functions
If g(x) = x2 + 1, f(x) = determine the compositions f(g(x)) and g(f(x))
Solution
We have f(g(x)) = f(x2 + 1) = =
On the other hand g(f(x)) = g+ 1 = 2 + 1 = + 1 =
=
Note that f(g(x)) g(f(x) of course.
3.3.6Inequalities
Find the ranges of values of x for which the following are satisfied
i)2x – 3 > 2ii)< 1
iii)x2 – x + 1 3iv)x2 + 2x + 2 < 5
v)|2x – 1| 2vi)< 4
vii)< – 3
Solution
Inequalities are difficult things to deal with in general, so we take them very steadily, checking each step. What you must not do is replace the inequality by an equals sign - we are not solving equations here and there is very rarely any need to do so.
i) If 2x – 3 > 2 then we have to rearrange to obtain x on its own. But we must be sure that what we are doing respects the inequality sign at each step. We have
2x – 3 > 2
so
2x > 3 + 2 = 5
Next, since 2 is positive we can divide both sides of the inequality by it without reversing the inequality sign to get
x >
which is the required answer.
ii) In solving something like < 1 we have to be very careful. We will want to multiply across by x – 3, but this will only leave the inequality unchanged if it is positive, ie if x – 3 > 0. If it is negative, ie when x – 3 < 0 then multiplying both sides by it will reverse the inequality. You may be tempted to square both sides to make everything positive, but this has its own dangers because you may be introducing new solutions. For example, think what happens if you square both sides of x = 1 (so ‘square with care’!). The safest thing to do is consider the different possible cases separately - and take it steady!
So, first consider x – 3 > 0. Then:
< 1 implies 2x < x – 3
or, subtracting x from both sides
x < – 3
But this is of course impossible if x – 3 > 0, so there are no solutions in this case. So now consider the case when x – 3 < 0. Then we are multiplying both sides of the inequality by a negative number and this reverses the direction of the inequality, so we get in this case
2x > x – 3 or x > – 3
So in this case we get a solution - we have x > – 3, and putting this with x – 3 < 0 we get
– 3 < x < 3
If you found this question difficult there is a good reason - it is difficult! We have to keep a lot of things in mind at once. The antidote is to do lots of examples!
iii) We can deal with x2 – x + 1 3 in a number of ways, but first it is better to rearrange it to give x2 – x – 2 0. Factorizing we then have
x2 – x – 2 = (x + 1)(x – 2) 0
We can take care of the equality first - the left-hand side is zero when x = – 1 or 2. Now (x + 1)(x – 2) will be positive if both factors are the same sign, so either x + 1 > 0 and x – 2 > 0, in which case x > 2, or x + 1 < 0 and x – 2 < 0 in which case x < – 1. So, putting these together, x2 – x – 2 0 if x – 1 or x 2.
Another approach is to complete the square (UEM 66) to get
x2 – x – 2 ∫2 – 0
so
2
Taking the square root this is equivalent to
x – or x – –
or, adding throughout
x – 1 or x 2
Yet another approach is to plot the graph of the function, as we did in RE 3.3.2B. Then it is 'obvious' that the quadratic x2 – x – 2 0 where the curve lies on or above the x-axis, ie where x 2 or x – 1. This is not however as 'rigorous' as the approaches used above.
iv)We can treat x2 + 2x + 2 < 5 as in the previous example. Rearranging it as x2 + 2x – 3 < 0 we have (x – 1)(x + 3) < 0, which can only occur if x – 1 and x + 3 are of opposite sign. So either
x – 1 < 0 and x + 3 > 0 , ie – 3 < x < 1
or
x – 1 > 0 and x + 3 < 0, which is impossible
So the only possibility is
– 3 < x < 1
Or again, you could plot the graph of the function and solve the identity by inspecting this.
v) Remember that |x| a is equivalent to – a x a and so |2x – 1| 2 is equivalent to
– 2 2x – 1 2
or, by now familiar arguments
– 2 + 1 = –1 2x 2 + 1 = 3
or, dividing through by 2:
– x
vi) Since |x – 4| is positive, we can multiply through by it without changing the direction of the inequality and thereby obtain from < 4, cancelling a 2 from both sides at the same time
1 < 2|x – 4| or < |x – 4|
This is equivalent to
– > x – 4 or x – 4 >
Adding 4 throughout now gives
x < or x >
vii) In the case of < – 3 it is safest to treat two cases separately, x – 1 > 0 and x – 1 < 0. Clearly, we cannot have x – 1 = 0. It is tempting in such circumstances to square both sides to obtain only positive quantities and you may care to go that way. But, as noted earlier, you should always 'square with care', since you may introduce new solutions that are not solutions of the original 'unsquared' equation.
If x – 1 > 0, then multiplying through by it won't change the direction of the inequality and we obtain 1 < – 3(x – 1) = – 3x + 3. You should now have no trouble in finding x < , which of course contradicts x > 1, and so is impossible. So we can only have x – 1 < 0 and the inequality reverses on multiplying through by it to give 1 > – 3(x – 1). From this we now find x > . Combining this with x – 1 < 0 gives finally
< x < 1
3.3.7Inverse of a function
Find the inverse functions for each of the following functions, specifying the values for which they exist.
i)2x + 1ii)x – 2
iii)x2 + 1, x 0
Solution
i) To find the inverse function f–1(x) of f(x) = 2x + 1 we write y = 2x + 1 and solve this for x in terms of y:
y = 2x + 1 implies 2x = y – 1 or x =
By definition this is f–1(y), so
f–1(y) =
or, expressing this in terms of x
f–1(x) =
This exists for all values of x
ii) Applying the same approach we have y = so (x + 2)y = x – 1, or rearranging
(y – 1)x = – 1 – 2y
or
x = = f–1(y)
and so
f–1(x) = provided x 1
iii)Putting y = x2 + 1 and solving for x gives x = = f–1(y) and so
f–1(x) = provided x 1
3.3.8Series and sigma notation
3.3.8A.
Write the following in sigma notation
i)1 + 8 + 27 + 64 + 125ii)3 + 6 + 9 + 12 + 15 + … + 99
iii)+ + + + … + iv)1 – + – + …
Solution
In all these problems the key is to find a general expression for the nth term in the series.
i) Notice that 1 + 8 + 27 + 64 + 125 = 13 + 23 + 33 + 43 + 53 and so the general nth term can be written n3 and we may therefore write the series as
n3
ii)3 + 6 + 9 + 12 + 15 + … + 99 = 3(1 + 2 + 3 + ... + 33) = 3n
iii)+ + + + … + =
iv) In 1 – + – + … the implication is that the series keeps on going to infinity, preserving the present pattern. The nth term in the sequence of denominators 1, 3, 9, 27 ... or 30 , 31 32, 33 ... is clearly 3n+1. The signs alternate in the form (– 1)n+1 and so the nth term is
(– 1)n+1
and the series may be written as a sum to infinity as
n+1
Note that alternative forms are possible, for example we could write iii) as or iv) as n .
B.Write down the first four terms in the series
i)ii)r!
iii)
Solution
i)= + + + = 1 + + +
ii)r! = 0! + 1! + 2! + 3! = 1 + 1 + 2 + 6
iii)= + + +
= + + +
3.3.9Finite series
Sum the geometric series:-
i)1 + + + + … + ii)n
iii)2niv)n
v)1 + 0.1 + 0.01 + 0.001 + 0.0001
Solution
i)In 1 + + + + … + the first term is a = 1 and the common ratio is r = . Noting that the first term is and the last is we are summing to six terms and so we have
1 + + + + … + = 1 = =
ii)n has a first term of 0.1 and common ratio of 0.1 and we are summing to 6 terms, so we have
n = 0.1 = 0.1 = 0.1 = 0.111111
iii) 2n = 2 = 2 (255) = 510
iv) n = = =
v) For 1 + 0.1 + 0.01 + 0.001 + 0.0001 you can use the sum formula if you wish, but you may prefer to notice that if you add 1 to ii) and subtract 0.000011 then you obtain the sum of the present series, which is thus 1.1111.
3.3.10Infinite series
3.3.10A.
Identify which are geometric sequences, and sum them to infinity.
i)1, 2, 3, 4, …ii)– 1, 3, 5, 9, …
iii)1, 1, 1, 1, …iv), , , , …
v)1, , , , …vi)0.1, 0.3, 0.5, 0.7, …
vii)2, – 4, 8, – 16, …viii)0.2, 0.04, 0.008, 0.0016, …
Solution
i) 1, 2, 3, 4, … is clearly not a geometric sequence since the successive terms are not all in the same ratio.
ii) Ditto for – 1, 3, 5, 9, …
iii) 1, 1, 1, 1, … is a geometric sequence, with first term 1 and common ratio 1, and is therefore divergent - it sums to infinity.
iv) , , , , … is not geometric, since the ratio of succesive terms varies.
v) 1, , , , … is geometric with first term 1 and common ratio . Since this is less than 1 the series converges to
=
vi) 0.1, 0.3, 0.5, 0.7, … is not geometric
vii) In 2, – 4, 8, – 16, … we see that successive terms are in the common ratio r = – 2, so this is a geometric series with first term 2 and common ratio – 2. Since r is of magnitude greater than 1 the series is divergent and cannot be summed.
viii) 0.2, 0.04, 0.008, 0.0016, … is a geometric series, with first term 0.2 and common ratio 0.2 so its sum is
= = 0.25
3.3.10B.
Find the sum of the infinite geometric series with first terms and common ratios given respectively by
i)1, 2ii)2,
iii)– 1, 1iv)1,
Solution
i) If first term is a = 1, and common ratio is r = 2 then this series is divergent and sums to infinity
ii) A series with first term a = 2 and common ratio r = < 1 has a sum to infinity of
= 4
iii) In this case the common ratio is 1 and the series is divergent to infinity.
iv) a = 1 and r = gives a sum of =
3.3.11Infinite binomial series
Find the first four terms of the binomial expansion of the following:
i)(1 + x)–1ii)(1 – 3x)–1
iii)(1 + 4x)–2iv)(1 – x)
Solution
i) To expand (1 + x)–1 we apply the binomial expansion
(1 + x)n = 1 + nx + ...+ x2 + x3 + …
+ xr + ...
We have
(1 + x)–1 = 1 + (– 1)x +
x2 + x3 + …
Take care with the signs. Tidying up the results we find for the first four terms
(1 + x)–1 = 1 – x + x2 – x3 + ...
since the factorials cancel on top and bottom.
ii) For (1 – 3x)–1 notice that the Review Question (3.1.11) is similar. We have
(1 – 3x)–1 = 1 + (– 1)(– 3x) +
2 + 3 + …
Tidying up the results we find for the first four terms
(1 + 3x)–2 =1 + 3x + 9x2 + 27x3 + ...
iii)(1 + 4x)–2 = 1 + (– 2)(4x) + 2 + 3 + …
Tidying up the results we find
(1 + 4x)–2 = 1 – 8x + 48x2 – 256x3 + ...
iv)(1 – x)= 1 + + + + ....
= 1 – – – – ...
– 1 –