Download Max/Min - UBC Math

Last week
Taylor approximations.
Taylor's remainder theorm.
This week... the moment you've all
been waiting for.
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Up until now most of the stuff I've been teaching
you has been abstract tools. Interesting, but not
always directly applicable (well... some of them
have some pretty sweet uses).
Today we actually get into the guts of what
Calculus is used for. Today we get to start doing
something useful with all the stuff you've
learned.
Today we optimize.
Optimise
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Okay, so what does optimize mean.
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According to google, Optimise means:
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“To make the best or most effective use of”
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This could mean the fastest, cheapest, tallest,
coolest, most efficient, most awesome, etc.
Whenever you see a word ending in “est” you're
probably dealing with some kind of optima.
Okay, great show me the numbers.
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Suppose we are trying to find the highest point
on a function.
Whenever the derivative of our function is
positive, we would be able to go somewhere
higher by increasing x by a small amount.
Whenever the derivative of our function is
negative, we could find our way to higher
ground by going backwards a small amount.
So... that leaves only only one possible
solution- the highest point must happen when
the derivative is zero.
Example
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“Find the minimum of (x+1)(x+5)”
Okay... so the derivative has to be zero where we are looking for.
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Find the derivative:
f'(x)= (x+1) + (x+5)
By the product rule
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f'(x)=0 = 2x + 6
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2x=-6 at minimum.
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Thus x=-3 at minimum.
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But wait... this still isn't the minimum value. To find that we have to stick our x
value back into our function to find the minimum function value.
(-3+1)(-3+5) = -4
Awesome.
Wait wait wait... hang on a minute.
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Okay, so, there's a problem with stuffs I just told
you. Can anyone spot it?
The rules we use to find the minimum and the maximum are exactly the
same!
We have no way of telling the difference between the two (or even weirder
objects)...
That's okay- I'm sure many of you have seen that before, and for those who
haven't, I'll deal with it next week. For now lets just continue.
Anywhere on a graph where the derivative of the function is zero, we refer to
as a critical points. Maxima and Minima are both different types of critical
points, but just because you find a critical point doesn't mean that you've
found the one you are looking for, but its a good place to look!
CLICKERS
4x
Where does the function f ( x)=e +e
have its minimum value?
ln 2
A) x=
6
ln 6
B) x=
2
−ln 2
C) x=
6
−ln 6
D) x=
2
−2x
CLICKERS
−x
What is the highest value of f ( x)= x e ?
A)1
B) 0.367 ..
C) 1/ √ 2
D) 0.463...
CLICKERS
3
What is the lowest value of f ( x)= x −3x ?
A)−2
B)−4
C) 4
D) 2
CLICKERS
3
What is the lowest value of f ( x)= x −3x ?
A)−2
B)−4
C) 4
D) 2
Not all functions
have a minimum.
Falls to negative
infinity.
CLICKERS
4
3
2
What is the lowest value of f ( x)=3x +4 x −12 x +9 ?
A)−23
B)−18
C) 4
D)−41
Bounded examples
2
What is the highest value of f ( x)=5+9x+3x −x
for -4<=x<=4?
Because we are blocked
In one direction at
Our end points, it is
Possible to have a
Extrema, even with no
Zero derivative.
3
Method for bounded examples
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Find all critical points (derivative =0).
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Find all end points.
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Find anywhere where derivative is undefined
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Compare each of them to determine which is
higher.
That is all you need.
Class question
x
What is the lowest value of f ( x)=tan ( x)−
2
for − π ≤ x≤ π ?
3
3
Class question
What is the lowest value of f ( x)=tan ( x)−2 x
for − π ≤ x≤ π ?
3
3
2
f ' ( x)=0 when sec ( x)−2=0
1
2
sec ( x)=2 ⇒ cos( x)=
√2
Familiar anyone?
Class question
What is the lowest value of f ( x)=tan ( x)−2 x
for − π ≤ x≤ π ?
3
3
2
f ' ( x)=0 when sec ( x)−2=0
1
2
sec ( x)=2 ⇒ cos( x)=
√2
Familiar anyone? f ' ( x)=0 when x= ±π
4
Need to check ± π AND ± π
4
3
Class question
What is the lowest value of f ( x)=tan ( x)−2 x
for − π ≤ x≤ π ?
3
3
2
f ' ( x)=0 when sec ( x)−2=0
1
2
sec ( x)=2 ⇒ cos( x)=
√2
Familiar anyone? f ' ( x)=0 when x= ±π
4
Need to check ± π AND ± π
4
3
Tan(pi/3)= sqrt(3)
tan(pi/4)=1
Pi/4
Pi/3
Class question
f (π /3)= √ 3−2 π/3≈−0.362344
f (π /4)=1−2 π /4≈−0.57079
f (−π/ 4)=−1+2 π /4≈0.57079
f (−π/3)=−√ 3 2 π/3≈0.362344
Tan(pi/3)= sqrt(3)
tan(pi/4)=1
Pi/4
Pi/3