Transcript Who wants to be a Physiology Millionaire?

Who wants
to be a
Millionaire?
Hosted by
Kenny, Katie, Josh and Mike
$100
What is the derivative of (3x2)1/2 ?
A - (1/2)(3x2)-1/2
B - (1/2)(6x)-1/2
1/2
C -- (6x)
c
(6x) // 2x(3)
2x(3)1/2
D - (2/3)(3x2)3/2
Explanation
Y = (3x2)1/2
d/dx[un] = nun-1u’
u = 3x2
n = 1/2
So, d/dx (3x2)1/2 = 1/2(3x2)-1/2(6x) =
(6x) / 2x(3)1/2
$200
How many critical numbers are
on the graph of 2x2(4x)
A-1
B-2
C-3
D-4
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.
$500
If mean value theorem applies, find all values of c
in the open interval (a,b) such that
f b  f a
f ' c  
f(x) = x2/3 [0,1]
ba

A - c = 7.431
B - c = 3.154
C - c = .296
D - Mean value does not apply
Explanation

Mean value applies because f(x) is
differentiable and continuous on the interval.
If
then f’(c) = 1.
f b  f a
f ' c  
b  a that the derivative must equal 1
this means
and that the values of x are the c values of the
function.
f’(c) = 1 at c = 2.96 on the interval.
$1,000
On what intervals is the graph
of f(x) = -8 / x3 increasing?
A - (-∞ , 0)
B - (0 , ∞)
C - f(x) is strictly decreasing
D - (-∞ , ∞)
Explanation
A function is increasing on all the intervals
that f’(x) > 0 [ f(x) has a slope that is
greater than 0]
From the graph of f’(x) you can see that for
-inf. < x < inf., x is > 0 and therefore on the
interval (-inf. , inf.) f(x) is increasing.
$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
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
$4,000
Determine the points of
inflection of the function
f(x) = (x3+2)(x4)
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)
Explanation
The x values of points of inflection on f(x)
exist where f’’(x) = 0.
When f(x) = (x3+2)(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)
$8,000
Find the limit as x app. Inf.
f(x) = 20/(x2 + 1)
A - Limit does not exist
B - Positive Infinity
C-0
D - Negative Infinity
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 /
1+x2=
20 
 2 
0
x 

0
x 2   1  1 0
 2   2 
x  x 
$16,000
On what intervals is the concavity
positive on
f(x) = -2x2(1-x2)
A - (-∞ , -0.707)
(0.707, ∞)
B - (-∞ , -0.408) (0.408, ∞)
C - (-0.408 , 0.408)
D - (-0.707 , 0.707)
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. >>
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 > 0
And that on (-.408 , .408) the concavity < 0
$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)πr3
A - ±2.639 in3
B - ±3.464 in3
C - ±3.464 in2
D - 2.639 in3
Explanation
First find the differential dv.
dv = 4πr2dr
Given is r = 5.25in and dr = ±0.01in so:
dv = 4π(5.25)2(±0.01) = ±3.464in3
$64,000
Non-graphing calculator
Find the slope and concavity at
x = -7 on y = (-2x3)(sin x)
A - m = -710.33
concavity > 0
B - m = 710.33 concavity < 0
C - m = 710.33 concavity > 0
D - m = -710.33 concavity < 0
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 >>
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.
$125,000
Non-graphing calculator
Find the concavity and the
equation of the tangent line at x
= 3 on y = (2)/(x2+x3)
A - conc <0 y = -0.051x + 0.209
B - conc >0 y = -0.051x
-1.051x + 0.209
C - conc >0 y = -2.675x - 3.081
D - conc >0 y = -2.675x + 3.081
Explanation
f(x) = (2)/(x2+x3) or (2)(x2+x3)-1
f(3) = .0556 = y
f’(x) = 2(-1(x2+x3)-2(2x +3x2))
f’(3) = -0.051 = m
f”(x) = -2((-2(x2+x3)-3)(2x +3x2) + ((x2+x3)-2)(2+6x)
f”(3) = 0.0625
f”(3) > 0 so the concavity is positive.
Use (y-y1)=m(x-x1) >>> (y - .0556 ) = (- .051)( x - 3 )
Simplify to get y = -0.051x + 0.209
$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
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.
$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 )
C-(0,0)
D - ( -1, 3890
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
When we plug -1, 0, and 2
7
3
f’(x) = 8x + 8x
into the f(x) we find that 2,
6
2
f”(x) = 56x + 24x
3680 is the absolute
5
f’’’(x) = 336x + 48x
maximum on the interval
f’’’(x) = 0 at x = 0
$1,000,000
The functions f and g are
differentiable for all real
numbers. The function of h is
given by h(x) = f(g(x)) - 6
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
What slope, h’(r), must exist on 1 < r < 3
A - m = 14
B - m = -5
C-m=5
D - m = -6
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