MA1M01 Calculus Assignment 4

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MA1M01 Calculus Assignment 4
MA1M01 Calculus Assignment 4
1. Differentiate the following:
(a)
f (x) = sin2 (x2 )
d
d
d
f (x) = 2 sin(x2 ) ·
sin(x2 ) = 2 sin(x2 ) cos(x2 ) x2
dx
dx
dx
= 2 sin(x2 ) cos(x2 )2x = 4x sin(x2 ) cos(x2 )
(b)
g(x) = cos(x3 ) sin(x2 + 3x)
d
d
d
g(x) = cos(x3 ) ·
sin(x2 + 3x) +
cos(x3 ) · sin(x2 + 3x)
dx
dx
dx
= cos(x3 ) cos(x2 + 3x) · (2x + 3) − sin(x3 ) · 3x2 · sin(x2 + 3x)
= (2x + 3) cos(x3 ) cos(x2 + 3x) − 3x2 sin(x3 ) sin(x2 + 3x)
(c)
h(x) = sin
(Note that h(x) = sin
g(x)
f (x)
cos(x3 ) sin(x2 + 3x)
sin2 (x2 )
)
1
g(x)
d g(x)
·
f (x)
dx f (x)
g(x)
f (x)g 0 (x) − f 0 (x)g(x)
= cos
·
f (x)
[f (x)]2
cos(x3 ) sin(x2 + 3x)
·
= cos
sin2 (x2 )
(sin2 (x2 ))(2x + 3)[cos(x3 ) cos(x2 + 3x) − 3x2 sin(x3 ) sin(x2 + 3x)]
sin4 (x2 )
4x sin(x2 ) cos(x2 )(cos(x3 ) sin(x2 + 3x))
−
sin4 (x2 )
d
h(x) = cos
dx
2. Consider the following function:
f (x) = 25x2 (x − 1)(x − 2)2 for − 0.1 < x < 2.1
(a) Find all critical points of f (x). (Hint: Differentiate using the produst rule and
factorise.)
The critical points occur when f 0 (x) = 0 (or when f 0 (x) doesn’t exist, but that
doesn’t happen here).
d
f (x) =
dx
=
=
=
50x(x − 1)(x − 2)2 + 25x2 (x − 2)2 + 50x2 (x − 1)(x − 2)
25x(x − 2)[2(x − 1)(x − 2) + x(x − 2) + 2x(x − 1)]
25x(x − 2)[2x2 − 6x + 4 + x2 − 2x + 2x2 − 2x]
25x(x − 2)[5x2 − 10x + 4]
Solve the qudratic:
x =
10 ±
√
10 ± 2 5
1
100 − 80
=
=1± √
10
10
5
√
So we have 4 critical points:
1
1
x = 0, 2, 1 − √ , 1 + √
5
5
(b) Which critical points are relative minima? Which are relative maxima?
By the previous part we have:
1
1
f 0 (x) = 25x(x − 2)(x − (1 − √ ))(x − (1 − √ ))
5
5
(i) for x < 0, all 4 factors are negative so f 0 (x) > 0
2
(ii) for 0 < x < 1 −
(iii) for 1 −
(iv) for 1 +
√1
5
√1
5
√1 ,
5
three of the factors are negative so f 0 (x) < 0
<x<1+
√1 ,
5
two of the factors are negative so f 0 (x) > 0
< x < 2, one of the factors are negative so f 0 (x) < 0
(v) for x > 2 all of the facrors are positive so f 0 (x) > 0
You could also determine the sign of f 0 (x) by picking any point on the above intervals and looking at the sign of f ’ on this point.
Using the first derivative test:
• (i)+(ii) and (iii)+(iv) tells us that x = 0 and x = 1 + √15 are relative maxima
• (ii)+(iii) and (iv)+(v) tells us that x = 1 − √15 and x = 2 are relative minima
(Alternatively we could have used the second derivative test.)
(c) Which of the relative minima is smallest? Which of the relative maxima is largest?
1
16
1
16
f (0) = 0, f (1 − √ ) = − √ ≈ −7.155, f (1 + √ ) = √ ≈ 7.155, f (2) = 0
5
5
5
5
So the largest relative maximum of f (x) is at x = 1 + √15 with a value
smallest relative minimum is at x = 1 − √15 with a value of − √165 .
3
16
√
and
5
the

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