Who wants to be a Millionaire?

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Who wants to be a Millionaire?

• Hosted by Kenny, Katie, Josh and Mike

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100
What is the derivative of (3x2)1/2 ?
A - (1/2)(3x2)-1/2
B - (1/2)(6x)-1/2
C - (6x) / 2x(3)1/2
c - (6x) / 2x(3)1/2
D - (2/3)(3x2)3/2
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Explanation
Y (3x2)1/2 d/dxun nun-1u u 3x2 n
1/2 So, d/dx (3x2)1/2 1/2(3x2)-1/2(6x)
(6x) / 2x(3)1/2
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200
How many critical numbers are on the graph of
2x2(4x)
A - 1
A - 1
B - 2
C - 3
D - 4
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Explanation
Critical numbers exist where the f(x) 0 f(x)
2x2(4x) or 8x3 So f(x) 24x2 24x2 0 Divide
each side by 24 to get x2 0 Square root each
side to find that x 0 and that there is only
one critical number.
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500
If mean value theorem applies, find all values
of c in the open interval (a,b) such that f(x)
x2/3 0,1
A - c 7.431
B - c 3.154
C - c .296
C - c .296
D - Mean value does not apply
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Explanation
Mean value applies because f(x) is differentiable
and continuous on the interval. If
then f(c) 1. this means that the
derivative must equal 1 and that the values of x
are the c values of the function. f(c) 1 at c
2.96 on the interval.
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1,000
On what intervals is the graph of f(x) -8 / x3
increasing?
A - (-8 , 0)
B - (0 , 8)
C - f(x) is strictly decreasing
D - (-8 , 8)
D - (-8 , 8)
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Explanation
A function is increasing on all the intervals
that f(x) gt 0 f(x) has a slope that is greater
than 0 From the graph of f(x) you can see that
for -inf. lt x lt inf., x is gt 0 and therefore on
the interval (-inf. , inf.) f(x) is increasing.
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2,000
What is the differential dy? Y (49 - x3)1/2
A - dy (1/2)(-3x2)-1/2dx
B - dy (1/2)(-3x2)(49 - x3)1/2dx
C - dy (1/2)(49 - x3)-1/2dx
D - dy (1/2)(-3x2)(49 - x3)-1/2dx
D - dy (1/2)(-3x2)(49 - x3)-1/2dx
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Explanation
To find the differentiable, derive, and then
multiply both sides by dx. Y (49 -
x3)1/2 Deriving you get dy/dx (49 -
x3)1/2 Multiply each side by dx to get dy
(49 - x3)1/2 dx
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4,000
Determine the points of inflection of the
function f(x) (x32)(x4)
A - (-0.830, 0.677)
A - (-0.830, 0.677)
B - (-1.046, 1.024)
C - (1.3 x 10-13 , 6 x 10-52)
D - (-0.760, 0.987)
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Explanation
The x values of points of inflection on f(x)
exist where f(x) 0. When f(x) (x32)(x4) or
x7 2x4, f(x) 7x6 8x3. So, f(x) 42x5
24x2 Using a graphing calculator find where
f(x) 0. For 0 42x5 24x2, x
-0.830. Now, substituting .83 into the original
equation, we find that he coordinate of the p of
I is (-0.830 , 0.677)
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8,000
Find the limit as x app. Inf. f(x) 20/(x2
1)
A - Limit does not exist
B - Positive Infinity
C - 0
C - 0
D - Negative Infinity
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Explanation
If you divide everything in an equation by x to
the highest power in the denominator, then plug
in infinity for x, you can find the limit as x
approaches infinity. Lim as x app. Infinity 20
/ 1x2
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16,000
On what intervals is the concavity positive on
f(x) -2x2(1-x2)
A - (-8 , -0.707) (0.707, 8)
B - (-8 , -0.408) (0.408, 8)
B - (-8 , -0.408) (0.408, 8)
C - (-0.408 , 0.408)
D - (-0.707 , 0.707)
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Explanation
The concavity of f(x) at any value of x is
determined by the sign ( or - ) of f(x). If
the sign is then the concavity is positive and
negative if the sign is -. Points of infection
divide intervals of different concavity. P of I
occur where f(x) 0 and f(x) 0 at x
0.408 so the intervals of different concavity
are (-inf. , -.408) , (-.408 , .408) , and (.408
, inf.) Explanation cont. gtgt
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Explanation cont.
By checking an x value in each interval in the
second derivative you find that on (-inf. ,
-.408) and (.408 , inf.) the concavity is gt 0
And that on (-.408 , .408) the concavity lt 0
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32,000
The radius of a ball measures 5.25 inches. If
the measurement is correct to within 0.01 inch,
estimate the propagated error in the volume of
the ball. V (4/3)pr3
A - 2.639 in3
B - 3.464 in3
B - 3.464 in3
C - 3.464 in2
D - 2.639 in3
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Explanation
First find the differential dv. dv
4pr2dr Given is r 5.25in and dr 0.01in
so dv 4p(5.25)2(0.01) 3.464in3
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64,000
Non-graphing calculator
Find the slope and concavity at x -7 on y
(-2x3)(sin x)
A - m -710.33 concavity gt 0
B - m 710.33 concavity lt 0
C - m 710.33 concavity gt 0
C - m 710.33 concavity gt 0
D - m -710.33 concavity lt 0
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Explanation
The slope at x -7 can be found by plugging -7
in for x in the derivative of the function. Use
the product rule to get f(x) (-6x2)(sin x)
(-2x3)(cos x) f(-7) 710.33
Explanation continued gtgt
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Explanation Cont.
Use the sign of f(-7) to find concavity f(x)
(-6x2)(sin x) (-2x3)(cos x) f(x) (-12x)(sin
x) (-6x2)(cos x) (-6x2)(cos x) (-2x3)(-sin
x) -47.789 Because the second
derivative at x -7 is negative, the concavity
at x -7 is negative.
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125,000
Non-graphing calculator
Find the concavity and the equation of the
tangent line at x 3 on y (2)/(x2x3)
A - conc lt0 y -0.051x 0.209
B - conc gt0 y -1.051x 0.209
B - conc gt0 y -0.051x 0.209
C - conc gt0 y -2.675x - 3.081
D - conc gt0 y -2.675x 3.081
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Explanation
f(x) (2)/(x2x3) or (2)(x2x3)-1 f(3) .0556
y f(x) 2(-1(x2x3)-2(2x 3x2)) f(3) -0.051
m f(x) -2((-2(x2x3)-3)(2x 3x2)
((x2x3)-2)(26x) f(3) 0.0625 f(3) gt 0 so
the concavity is positive. Use (y-y1)m(x-x1)
gtgtgt (y - .0556 ) (- .051)( x - 3 ) Simplify to
get y -0.051x 0.209
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250,000
A farmer plans to fence a rectangular pasture
adjacent to a river. The pasture must contain
180,000m2 in order to provide enough grass for
the herd. When looking for the dimensions that
requires the least amount of fencing, what
equation should be set to zero to solve for the
length of side across from the river?
A - y 2x 180,000/x
B - y x 360,000/x
C - y 2 - 180,000/x2
D - y 1 - 360,000/x2
D - y 1 - 360,000/x2
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Explanation
Primary F 2x y Secondary xy 180,000 We
know that y is the side across from the river
because there is only one of that side.
Therefore we want to be solving for y. x
180,000 / y. Substitute x into the primary. F
2(180,000)/y y Now, \set 1 - 360,000/y2 equal
to zero to find the minimum value for y.
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500,000
Non-graphing calculator
Find the absolute maximum of the function f(x)
on -1 , 2 f(x) x8 2x4
A - ( 2 , 4712 )
B - ( 2 , 3680 )
B - ( 2 , 3680 )
C - ( 0 , 0 )
D - ( -1, 3890
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Explanation
The absolute maximum of f(x) is determined by
testing all of the critical numbers and endpoints
of f(x). The critical numbers are determined by
setting f(x) equal to zero. f(x) x8
2x4 f(x) 8x7 8x3 f(x) 56x6 24x2
f(x) 336x5 48x f(x) 0 at x 0
When we plug -1, 0, and 2 into the f(x) we find
that 2, 3680 is the absolute maximum on the
interval
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1,000,000
x f(x) f(x) g(x) g(x)
1 6 4 2 5
2 9 2 3 1
3 10 -4 4 2
4 -1 3 6 7
The functions f and g are differentiable for all
real numbers. The function of h is given by h(x)
f(g(x)) - 6
What slope, h(r), must exist on 1 lt r lt 3
A - m 14
B - m -5
B - m -5
C - m 5
D - m -6
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Explanation
First find h(1) and h(3) and find the slope that
is created by the two points. Then, by the
definition of the mean value theorem, there must
be a point on the interval with that slope. h(x)
f(g(x)) - 6 h(1) f(g(1)) - 6 f(2) - 6 9
- 6 3 h(3) f(g(3)) - 6 f(4) -6 -1 - 6
-7 m (y2 - y1) / (x2 - x1) (-7 - 3) / (3 - 1)
-5