Download Exercises for Thursday and Friday

The University of Sydney
MATH1111 Introduction to Calculus
Semester 1
Week 5 Exercises (Thurs/Fri)
2017
Important Ideas and Useful Facts:
(i) Common Functions: The functions that you are most likely to meet are built from the
following list, and some you have not seen before will be explained and illustrated later:
(a) linear: y = ax + b , where a (the slope) and b (the y-intercept) are constants
(b) quadratic: y = ax2 + bx + c , where a, b, c are constants
(c) cubic: y = ax3 + bx2 + cx + d , where a, b, c, d are constants
(d) polynomial: y = a0 + a1 x + a2 x2 + . . . + an xn , where a0 , a1 , . . . , an are constants
√
(e) square root: y = x , the inverse of the square function y = x2
(f) nth root: y = x1/n , the inverse of the nth power function y = xn
p(x)
(g) rational: y =
where p and q are polynomial functions
q(x)
(h) simple exponential: y = ax , where a is a positive constant
(i) general exponential: y = Aekx , where A, k are constants and e is Euler’s number
(j) logarithmic: y = loga x , where a is a constant, called the base of the logarithm
(k) natural logarithmic: y = ln x = loge x , the inverse of the exponential function y = ex
sin x
(l) trigonometric: y = sin x , y = cos x and y = tan x =
cos x
x
−x
e −e
ex + e−x
(m) hyperbolic: y = sinh x =
and y = cosh x =
2
2
(
x if x ≥ 0 ,
(n) absolute value: y = |x| =
−x if x < 0 .
(ii) Composites: We can build new functions out of old ones using arithmetic operations and
composition: if f and g are functions then the composite function f ◦ g is defined by the
rule
y = f ◦ g (x) = f (g(x))
whenever this combined rule makes sense. (Some functions are not compatible for composition, or only compatible for certain inputs and outputs.)
(iii) Vertical Line Test: A curve in the xy-plane is the graph of some function f if and only
if it satisfies the vertical line test, which means that as a vertical line scans from left to
right, it only passes through the curve in at most one place. In this case, a potential
input on the x-axis can correpond to at most one potential output on the y-axis, which
then becomes the unique value f (x) of the proposed function.
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(iv) Horizontal Line Test and Inverse Functions: If a horizontal line scanning from top to bottom
only passes through the graph of a function y = f (x) in at most one place, then we say
that the graph satisfies the horizontal line test and that the function f is one-one. This
means that each potential output on the y-axis can correspond to at most one input on
the x-axis.
Thus if we reflect the graph in the line y = x (interchanging the roles of x and y), then
the new graph satisfies the vertical line test, and we get a new function, called the inverse
of f , denoted by f −1 . The rule for f −1 “undoes” or “reverses” the rule for f :
y = f (x)
if and only if
x = f −1 (y) .
A function whose graph does not satisfy the horizontal line test may have its domain
restricted so that it does become one-one. For example, f (x) = x2 is defined for all real
x and is not one-one. But if we restrict the domain
√ to [0,−1∞) or (−∞,
√ 0] then f becomes
−1
−1
one-one with inverse f and the rule f (x) = x or f (x) = − x respectively.
The roles of domain and range are exchanged as one passes from a one-one function f to
its inverse f −1 . It is standard to use the symbol x for typical inputs, so the rules become
y = f (x) and y = f −1 (x) respectively.
(v) Powers and nth
roots form an inverse pair: If n is a positive integer then y = xn and
√
1/n
n
y=x
= x are mutually inverse functions, for all real x when n is odd,√and for all
nonnegative real x when n is even. For example, y = x2 and y√ = x1/2 = x form an
inverse pair for nonnegative reals, and y = x3 and y = x1/3 = 3 x form an inverse pair
for all reals.
(vi) Polynomials: A polynomial has the form
p(x) = a0 + a1 x + a2 x2 + . . . + an xn
where n is a nonnegative integer and a0 , . . . an are constants, with an 6= 0. We call n the
degree of the polynomial. For example, quadratics have degree 2 and cubic polynomials
have degree 3. A root of p(x) is a (real) number α such that p(α) = 0, where p(α) is
evaluated in R using addition and multiplication. If α is a root of p(x) then we have a
factorisation
p(x) = (x − α)q(x)
where q(x) is some polynomial of degree n − 1.
(vii) Division of polynomials: If p(x) and s(x) are polynomials, we can divide p(x) by s(x) to
get an equation
p(x) = s(x)q(x) + r(x)
for some polynomials q(x), called the quotient, and r(x), called the remainder, where
either r(x) = 0 or the degree of r(x) is less than the degree of s(x). In the case where
r(x) = 0 then the previous equation becomes a factorisation p(x) = s(x)q(x) .
(viii) Complete factorisation of polynomials: If p(x) is a polynomial of degree n and
p(x) = λ(x − α1 )(x − α2 ) . . . (x − αn )
for some constant λ and roots α1 , α2 , . . . , αn (possibly with repetition), then the previous
equation is called a complete (linear) factorisation of p(x).
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Tutorial Exercises:
1.
Let p(x) = x3 + 6x2 + 11x + 6. Evaluate p(0), p(1), p(2), p(3), p(−1), p(−2), p(−3). You
should discover three distinct roots. Now write down a complete factorisation of p(x).
2.
Which of the following graphs represent functions? one-one functions?
3.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Find the rules for f ◦ g, g ◦ f and g −1 when
f (x) = x2 + 6
and
g(x) =
√
2x − 7 .
Find two rules for f −1 by restricting the domain of f to [0, ∞) and to (−∞, 0].
4.
An observer on a flat plain views a rocket launch 2 kilometres away.
(i) Express the height h of the rocket in metres as a function of the angle θ of elevation
from the viewpoint of the observer.
(ii) Find the angle of elevation to the nearest degree when h = 10, 000.
∗
5.
A 50 metre pole casts a shadow on a flat plain.
(i) Express the angle θ of elevation of the sun in radians as a function of the length s
in metres of the pole’s shadow.
(ii) Find the angle of elevation to the nearest degree when s = 20.
∗
6.
Find the quotient of x3 − 1 by x − 1, the quotient of x3 + 1 by x + 1 and the quotients of
x4 − 1 by x − 1 and x + 1 respectively. Without doing any long division of polynomials,
why does x − 1 not divide x3 + 1? Why does x + 1 does not divide either x3 − 1 or x4 + 1?
Further Exercises:
7.
Find the quotient q(x) and the remainder r(x) in each case, when p(x) is divided by s(x):
(i) p(x) = x2 + 3x + 4 , s(x) = x + 1
(ii) p(x) = x2 − 3x − 4 , s(x) = x + 1
(iii) p(x) = 3x3 − 4x − 1 , s(x) = x − 2
(iv) p(x) = 3x3 − 4x − 1 , s(x) = x2 + 1
(v) p(x) = x4 + 3x3 + x2 − 5x + 10 , s(x) = x2 − x + 2
3
∗
8.
A 200 metre rope anchors a hot-air balloon floating over a horizontal plain. Assume the
rope is approximately straight and makes an angle θ radians with the plain.
(i) Express θ as a function of the height h (in metres) of the balloon.
(ii) Find the angle of elevation to the nearest degree when h = 150.
∗
9.
Solve for x for each of the following equations:
(i) x3 + 3x2 + 4x + 12 = 0 , given that −3 is a root.
(ii) 3x4 + 14x3 + 14x2 − 8x − 8 = 0 , given that −2 and − 32 are roots.
∗∗
10.
Graph the following functions:
(i) y = sin−1 x (ii) y = cos−1 x (iii) y = − cos−1 x (iv) y =
π
2
− cos−1 x
Explain the connection with the trig identity cos θ = sin(θ + π2 ).
Short Answers to Selected Exercises:
1.
6, 24, 60, 120, 0, 0, 0 ; roots are −1, −2, −3 ; complete factorisation is (x+1)(x+2)(x+3).
2.
3.
(ii), (iii) and (vi) represent functions, and (vi) represents a function that is one-one.
√
√
2
f ◦ g (x) = 2x − 1 , g ◦ f (x) = 2x2 + 5 , g −1 (x) = x 2+7 , f −1 (x) = x − 6 when the
√
domain of f is [0, ∞), and f −1 (x) = − x − 6 when the domain of f is (−∞, 0] .
4.
(i) h = 2000 tan θ (ii) 79 degrees
5.
(i) θ = tan−1 (50/s) (ii) 68 degrees
6.
x2 + x + 1 , x2 − x + 1 , x3 + x2 + x + 1 , x3 − x2 + x − 1
7.
(i) x + 2, 2 (ii) x − 4, 0 (iii) 3x2 + 6x + 8, 15 (iv) 3x, −7x − 1
(v) x2 + 4x + 3, −10x + 4
8.
(i) θ = sin−1 (h/200) (ii) 49 degrees
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