Download Quiz 12 - FAU Math

Department of Mathematical Sciences
Instructor: Daiva Pucinskaite
Calculus I
July 13, 2016
Quiz 12
Functions from derivatives. Use the derivative f ′ given by
f ′ (x) = x2 (x + 2)(x − 1)
to determine the x-coordinates of the local maxima and minima of f , and the intervals of increase and decrease. Sketch a possible graph of f (f is not unique).
Since f ′ (x) does not fail to exist for all x, the critical points of f are
x = −2, x = 0, x = 1
because at those three points f ′ is 0.
• f ′ (−3) = (−3)2 (−3 + 2) (−3 − 1) > 0 ⇒ f increases on (−∞, −2),
| {z } | {z } | {z }
>0
<0
<0
• f ′ (−1) = (−1)2 (−1 + 2) (−1 − 1) < 0 ⇒ f decreases on (−2, 0),
| {z } | {z } | {z }
>0
>0
<0
• f ′ (0.5) = (0.5)2 (0.5 + 2) (0.5 − 1) < 0 ⇒ f decreases on (0, 1),
| {z } | {z } | {z }
>0
>0
<0
• f ′ (5) = (5)2 (5 + 2) (5 − 1) > 0 ⇒ f increases on (−1, ∞),
|{z} | {z } | {z }
>0
>0
>0
Recall: Suppose c is a critical point of f , then
∗ f has a local maximum at c if f ′ changes sign form positive to negative, as x increases through.
∗ f has a local minimum at c if f ′ changes sign form negative to positive, as x increases through.
• Since f increases on (−∞, −2), and decreases on (−2, 0), i.e. f ′ changes sign form positive
to negative, the function f has a local maximum at the critical point c = −2.
• Since f decreases on (0, 1), and increases on (−1, ∞), i.e. f ′ changes sign form negative
to positive, the function f has a local minimum at the critical point c = 1.
• Since f decreases on (−2, 1), thus f ′ does not changes sign at the critical point c = 0. The
function f has neither a local minimum nor a local maximum at c = 0.
A possible graph of f .
3
2
1
-3
-2
-1
0
-1
1
2
3
Department of Mathematical Sciences
Instructor: Daiva Pucinskaite
Calculus I
July 13, 2016
Quiz 12
Maximum product. Find numbers x and y satisfying the equation 3x + y = 12 such
that the product of x and y is as large as possible.
Two numbers x and y satisfy the equation 3x + y = 12 if y = 12 − 3x. Thus we are
searching for a number x such that
x · y = x · (12 − 3x) is as large as possible.
If the function
f (x) = x(12 − 3x) = 12x − 3x2
takes the absolute maximum at a, i.e.
f (a) ≥ f (x) for all x,
|{z}
|{z}
a(12−3a)
x(12−3x)
then for x = a and y = 12 − 3a we have
• 3x + y = 12, and
• xy ≥ xy for all x and y with 3x + y = 12.
Since f ′ (x) = (12x − 3x2 )′ = 12 − 6x implies that
f ′ (c) = 12 − 6c = 0 for c = 2,
the function f has only one critical point c = 2. Because f ′ (1) = 12 − 6 · 1 > 0 and f ′ (3) =
12 − 6 · 3 < 0, the function f has a local maximum at 2. Since 2 is the only critical point, the
function f takes at the local maximum the absolute maximum. We have f (2) ≥ f (x) for all x.
Therefore for
x = 2 and y = 12 − 3 · 2 = 6
we have that the product 2 · 6 = 12 is larges number among xy with 3x + y = 12.