To help you get started, here is #1 on page192:
f ( x)  5 x  x 2
a) Find the local extrema
f ( x)  5  2 x  0
This point occurs at x 
5
2
Notice also that f (x) is an upside down parabola. So this point is a max.
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














To help you get started, here is #1 on page192:
f ( x)  5 x  x 2
a) Find the local extrema
f ( x)  5  2 x  0
This point occurs at x 
b) Find the intervals on which the function is increasing
5
2
and
c) Find the intervals on which the function is decreasing
If we hadn’t already noticed that it is a parabola, we could have just
tested the intervals of the derivative:
f ( x)  5  2 x  0
f ( x)  5  2 x  0
5
when x 
2
5
when x 
2
So f is increasing
So f is decreasing
#15 on page 192
f ( x)  x 2  2 x  1
on the interval [0, 1]
Show that the function f satisfies the hypotheses of the Mean Value
Theorem on the given interval [a, b]
We start by looking at this graph (again a parabola) on the interval [0, 1]
It is continuous (no holes or breaks in
the graph) and differentiable (no cusps
or corner points) on the interval [0, 1]





That’s all that it takes to satisfy the
hypotheses of the MVT.
#15 on page 192
f ( x)  x 2  2 x  1
on the interval [0, 1]
Find each value of c that satisfies the Mean Value Theorem on the given
interval [a, b]
We start by drawing a secant line over the interval [0, 1]
This satisfies:





f (b)  f ( a )
ba
2  (1)

3
1 0
So the slope of the secant line is 3.

f (b)  f (a )

f
(
c
)

Now we have to find a number c between 0 and 1 such that…

ba
#15 on page 192
f ( x)  x 2  2 x  1
on the interval [0, 1]
Find each value of c that satisfies the Mean Value Theorem on the given
interval [a, b]
We start by drawing a secant line over the interval [0, 1]
Now we have to find a number c
between 0 and 1 such that…



f (c) 




f (c)  2c  2 
f (b)  f (a )
ba
In this case, we know that the
right side is equal to 3. So…
The derivative of f at x = c in this case is...
f (1)  f (0)
3
1 0
Solving gives us c =
1
2
#15 on page 192
f ( x)  x 2  2 x  1
on the interval [0, 1]
Find each value of c that satisfies the Mean Value Theorem on the given
interval [a, b]
We start by drawing a secant line over the interval [0, 1]



This means that at x = ½, the
tangent line is parallel to the secant
line that we’ve already drawn.





f (b)  f (a )
f (c) 
ba
Let’s enlarge the graph so we can get a better look at the two parallel lines




f (c) 

f (b)  f (a )
ba


1
c
2
