Download Quiz 6 Solutions

Math 134 Fall 2011 Quiz 6
November 7, 2011
Name:
1. Consider the function f (x) = x3 e−x on [−1, 4].
(a) Find all critical values of f (x). Show your work.
Critical values are values of x where f 0 (x) is zero or undefined. Here f 0 (x) = 3x2 e−x −
x3 e−x = (3x2 − x3 )e−x . Since f 0 (x) is defined everywhere there are no places where
the derivative is undefined. Since e− x is always positive, the derivative is equal to
zero when
(3x2 − x3 ) = 0
or
x2 (3 − x) = 0.
Thus there are critical values at x = 0 and x = 3.
(b) Find the global minimum and maximum values of f (x) on the interval [−1, 4].
Show all your work.
First we notice that this function is continuous on its domain so there must be a
global maximum and minimum on this interval. So we must test the critical values
and end points in the original function.
x
f (x)
3 −(−1)
-1
(−1) e
= −e ≈ −2.71828
0
f (0) = 03 e0 = 0
3 f (3) = 33 e− 3 = 27e−3 ≈ 1.34425
4 f (4) = 43 e−4 = 64e−4 ≈ 1.172200
Our conclusion is that there is a global maximum of 27e−3 occurring when x = 3 an
there is a global minimum of −e occurring when x = −1.
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2. Consider the function g(x) = x2 −x+1 on [1, 4]. State why g(x) satisfies the hypotheses
of the Mean Value Theorem on this interval, and then find all values of c guaranteed
in the conclusion of the Mean Value Theorem.
The function g(x) is a polynomial so it is continuous and differentiable for all real
numbers and thus is continuous on [1, 4] and is differentiable on (1, 4).
.
The Mean Value Theorem states that there is a value c such that g 0 (c) = g(4)−g(1)
4−1
0
Computing g (x) = 2x − 1 we have that:
2c − 1 =
(16 − 4 + 1) − (1 − 1 + 1)
=4
3
or
2c − 1 = 4
or c = 5/2.
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3. Using a linear approximation at a “nice” value, approximate (63.76) 3 .
1
In this problem the function we want to use is f (x) = x 3 . The “nice” value is a = 64
since f (64) = 4. We note that f 0 (x) = 12 . Thus our linear approximation at 8 is
3x 3
given by:
L(x) = f (64) + f 0 (64)(x − 64)
or
L(x) = 4 +
1
2
3(64 3 )
or
L(x) = 64 +
(x − 64)
1
(x − 64)
48
Now we evaluate at x = 63.76
L(63.76) = 64 +
1
1
1
(63.76 − 64) = 64 + (−.24) = 64 −
= 63.995
48
48
200
2
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4. Consider the function f (x)= x 3 (7 − x2 ). Find the global maximum and minimum of
f (x) on the interval −2, 21 .
First we identify the critical values of f (x), by finding where the derivative is equal
to 0 or undefined. We use the product rule to compute f 0 (x).
1
1 −2
x 3 (7 − x2 ) + x 3 (−2x))
3
4
4
7
x3
=
− 2x 3
2 −
3
3x 3
4
7
7x 3
=
2 −
3
3x 3
7
1 − x2
=
2
3x 3
f 0 (x) =
Looking at the derivative we see that the derivative is undefined if x = 0. The
derivative is 0 if 1 − x2 = 0, solving this gives x = ±1. Thus the critical values of
f (x) are x = 0 and x = ±1. However 1 is not in the interval [−2, 21 ] so we do not
need to consider this value. We the test the function f (x) at each critical point in
our interval and at the endpoints of the interval
f (−2) = (−2)1/3 (7 − 4) ≈ −3.779
f (−1) = −1(6) = −6
f (0) = 0
13 1
1
f (1/2) =
7−
≈ 5.35748
2
4
The largest value obtained is 5.35748 and thus f (x) has a global maximum of occurring at x = 21 . The smallest value is −6, thus f (x) has an absolute minimum of −6
occurring at x = −1.
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