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- Critical numbers occur when the derivative is zero or undefined
3/ 3 2 / 3 5 / 3 5 / 3 5/ 3 5/ 3
2 2 / 3 2 / 3 2 / 3
f x x x
x f x x x x x x x x
x x f x x x x x x
− −
- The function is continuous on the interval given [-1,2]
2 2 2 2 2
2 2 2 2 2
2 2
2
: 4 0 2, and 2 0 2
x x x x x f x x x
x x x x x
CP x x x x
f
f
f
The absolute max is at sqrt(2), the absolute min is at -1. Note we throw out the other points because
they are outside our interval.
Math 181
Chapter 4 Review
- f is continuous on [0,2] and differentiable on (0,2)
f (0) = -1 and f (2) = 9. There is a c in (0,2) where
2 2
f c
f x x x x
c
2 ( ) 6 3 3 ( 2)
f x x x x x
f x x x
increasing for x < 0 and x > 2
decreasing for x between 0 and 2
concave up for x > 1
concave down for x < 1
critical points at 0 and 2
inflection point at 1
- For the function
sin ( ) 1 cos
x f x x
a) Find the first and second derivatives
2
2 2
2 2
(1 cos ) cos sin cos 1 1 ( ) (1 cos ) (1 cos ) 1 cos
sin ( ) 1(1 cos ) ( sin ) (1 cos )
x x x x f x x x x
x f x x x x
−
b) The first derivative is undefined when cosine is negative 1, or ±(2 k + 1) π for k = 0,1,2…
The first derivative is never zero and is always positive (when defined)
The second derivative is undefined at the same place as above ± (2 k +1) π
The second derivative is zero when sine is zero, or ± 2 k π
c) Critical points: ± (2 k +1) π
Inflection points: ± 2 k π
d)
- If 1000 sq cm of material is available to make a box with a square base and an open top, find the
largest possible volume of the box
Box dimensions: base x , height h
Volume of box:
2 x h
Surface area of box: bottom + 4 sides =
2 x + 4 xh ≡ 1000. So
2 1000
x h x
Volume =
2 2 1000 1 3 (1000 ) 4 4
x x x x x
Differentiating we find
− x ≡ x =
The second derivative is always negative, so this is indeed a max
2 1000 18.
4(18.26)
h
To maximize the volume, the box should have a base of 18.26 cm and a height of 9.13cm
2 2 23 ( ) ( ) ( ) 10 (650 3 0.05 ) 7 650 210 420
x x P x R x C x x x x x
Differentiating we find
x P ′^ x = − ≡ ⇒ x =
This is a max, because the second derivative is always negative (concave down)
4 3 2 3 2 ( x + 4 x − 7 x − 20 x + 23) ′= 4 x + 12 x − 14 x − 20
f x f x x f x f
. f (1.90625) =0.
f x f x x f x f
= − = − =. f (1.892015) =0.
So the root is approximately 1.
4 3 2 4 3 7 2 ( ) 20 12 7 4 5 4 4 4 3 2 2
x x x g x = + + + x + c = x + x + x + x + c
g (2) = 1 implies that
g c
c c
So
g x = x + x + x + x −
