# 10350 Tutorial Week 10 - Set 01 Solutions

**10350 Tutorial Week 10 - Set 01**

1a. Find all critical points of

** Solution .** We have
This is undefined for x = 1, and we have

gives (x -1)^{2/3} = 4, so (x -1)^{1/3} = ±2, thus x = -7, 9,
thus our critical points are at x = -7, 1, 9.

1b. Find the absolute maximum and absolute minimum of
for -7 ≤ x ≤ 28.

** Solution .**

This gives our absolute maximum of 5/4 at x = 9, and the absolute minimum of
-17/12 at x = -7 (note

that all four points to check are in the domain -7 ≤ x ≤ 28).

**10350 Tutorial Week 10 - Set 02
**1a. Find all critical points of

**Solution .**, which is defined for all x.

Then so so 1 - 4x

^{2}= 0, so x

^{2}= 1/4, so x = ±1/2 are the critical

points of g(x).

1b. Find the absolute maximum and absolute minimum of
for 0 ≤ x ≤ 1.

** Solution .** f(0) = 0

This gives the absolute maximum of
at x = 1/2 and the absolute minimum of 0 at x = 0. Note

that we don't consider x = -1/2 since it is not in the given domain of 0 ≤ x ≤
1.

**10350 Tutorial Week 10 - Set 03 **

1. At noon, ship A is 10 km due East of ship B. Ship A is sailing west at 3
km/hr, and ship B is sailing

south at 1 km/hr. At what time will the ships be nearest to each other, what
will this distance be?

** Solution . **We can assume that ship B is at the
origin at time 0, and ship B is then 10 miles east of

the origin. Note that they start out 10 km from each other (this is the endpoint
for t = 0). We know

that at time t, we have ship B is t units south of the origin and ship A is 10 -
3t units from the origin.

If c is the distance between them, we have (from the Pythagorean Theorem) c^{2} =
(10 - 3t)^{2} + t^{2}, or

so by the chain rule
The function is undefined when

the denominator is zero , so when 100 - 60t + 10t^{2} = 10(10 - 6t + t^{2}) = 0, or
10 - 6t + t^{2} = 0, but the

discriminant (the part under the square root in the quadratic formula, b^{2}-4ac)
is (-6)^{2}-4(1)(10) = -4,

hence the quadratic formula gives imaginary answers for x, so the function is
always defined. Thus, we

only need to set c' = 0, giving that the numerator must be 0, so we get 20t - 60
= 0, or t = 3. Since,

at t = 3 our distance is
which is less that the distance of 10 at t = 0,

we know our minimum occurs at 3PM with a distance of
km.

**10350 Tutorial Week 10 - Set 04**

1. Consider the function f(x) = x ln(x^{2}).

1a. Find the derivative of f(x).

** Solution .** f'(x) = ln(x^{2}) + 2

1b. Using Q1(a), verify that the function f(x) = x ln(x^{2})
satisfies the hypotheses of the Mean Value

Theorem on [1, e]. Explain clearly in words.

** Solution . **Our function f is continuous on [1, e],
and the derivative exists on the interval (1, e).

Thus we can apply the MVT.

1c. Find all numbers c that satisfies the conclusion of the
Mean Value Theorem for f(x) = x ln(x^{2}) .

** Solution . **We need c with
so

Thus,
so
giving the answer (in the domain: [1, e]) of

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