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$100$100

What is the derivative of (3x2)1/2 ?

B - (1/2)(6x)-1/2

A - (1/2)(3x2)-1/2

C - (6x) / 2x(3)1/2

D - (2/3)(3x2)3/2

c - (6x) / 2x(3)1/2

ExplanationExplanation

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$200

How many critical numbers are on the graph of 2x2(4x)

B - 2

A - 1

C - 3

D - 4

A - 1

ExplanationExplanation

Critical numbers exist where the f’(x) = 0

f(x) = 2x2(4x) or 8x3

So f’(x) = 24x2

24x2 = 0Divide each side by 24 to get x2 = 0Square root each side to find that x = 0 and that there is only one critical number.

$500$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]

B - c = 3.154

A - c = 7.431

C - c = .296

D - Mean value does not apply

C - c = .296

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f ' c( ) =f b( ) − f a( )b− a

ExplanationExplanation

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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f ' c( ) =f b( ) − f a( )b− a

$1,000$1,000

On what intervals is the graph of f(x) = -8 / x3 increasing?

B - (0 , ∞)

A - (-∞ , 0)

C - f(x) is strictly decreasing

D - (-∞ , ∞)D - (-∞ , ∞)

ExplanationExplanationA 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$2,000

What is the differential dy?Y = (49 - x3)1/2

B - dy = (1/2)(-3x2)(49 - x3)1/2dx

A - dy = (1/2)(-3x2)-1/2dx

C - dy = (1/2)(49 - x3)-1/2dx

D - dy = (1/2)(-3x2)(49 - x3)-1/2dxD - dy = (1/2)(-3x2)(49 - x3)-1/2dx

ExplanationExplanation

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$4,000

Determine the points of inflection of the function f(x) = (x3+2)(x4)

B - (-1.046, 1.024)

A - (-0.830, 0.677)

C - (1.3 x 10-13 , 6 x 10-52)

D - (-0.760, 0.987)

A - (-0.830, 0.677)

ExplanationExplanationThe 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$8,000

Find the limit as x app. Inf. f(x) = 20/(x2 + 1)

B - Positive Infinity

A - Limit does not exist

C - 0

D - Negative Infinity

C - 0

ExplanationExplanation

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=

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20

x 2

⎛

⎝ ⎜

⎞

⎠ ⎟

x 2

x 2

⎛

⎝ ⎜

⎞

⎠ ⎟+

1

x 2

⎛

⎝ ⎜

⎞

⎠ ⎟

=0

1+ 0= 0

$16,000$16,000

On what intervals is the concavity positive on

f(x) = -2x2(1-x2)

B - (-∞ , -0.408) (0.408, ∞)

A - (-∞ , -0.707) (0.707, ∞)

C - (-0.408 , 0.408)

D - (-0.707 , 0.707)

B - (-∞ , -0.408) (0.408, ∞)

ExplanationExplanation

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 Explanation cont.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$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

B - ±3.464 in3

A - ±2.639 in3

C - ±3.464 in2

D - 2.639 in3

B - ±3.464 in3

ExplanationExplanation

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$64,000

Find the slope and concavity at x = -7 on y = (-2x3)(sin x)

B - m = 710.33 concavity < 0

A - m = -710.33 concavity > 0

C - m = 710.33 concavity > 0

D - m = -710.33 concavity < 0

C - m = 710.33 concavity > 0

Non-graphing calculator

ExplanationExplanation

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 getf’(x) = (-6x2)(sin x) + (-2x3)(cos x)f’(-7) = 710.33

Explanation continued >>

Explanation Explanation Cont.Cont.

Use the sign of f’’(-7) to find concavityf’(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$125,000

Find the concavity and the equation of the tangent line at x = 3 on y = (2)/(x2+x3)

B - conc >0 y = -1.051x + 0.209

A - conc <0 y = -0.051x + 0.209

C - conc >0 y = -2.675x - 3.081

D - conc >0 y = -2.675x + 3.081

B - conc >0 y = -0.051x + 0.209

Non-graphing calculator

ExplanationExplanation

f(x) = (2)/(x2+x3) or (2)(x2+x3)-1

f(3) = .0556 = yf’(x) = 2(-1(x2+x3)-2(2x +3x2))f’(3) = -0.051 = mf”(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$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?

B - y = x +360,000/x

A - y = 2x + 180,000/x

C - y = 2 - 180,000/x2

D - y = 1 - 360,000/x2D - y = 1 - 360,000/x2

ExplanationExplanation

Primary: F = 2x + ySecondary: 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 + yNow, \set 1 - 360,000/y2 equal to zero to find the minimum value for y.

$500,000$500,000

Find the absolute maximum of the function f’’(x) on [-1 , 2] f(x) = x8 + 2x4

B - ( 2 , 3680 )

A - ( 2 , 4712 )

C - ( 0 , 0 )

D - ( -1, 3890

B - ( 2 , 3680 )

Non-graphing calculator

ExplanationExplanation

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 + 48xf’’’(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

$1,000,000$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

B - m = -5

A - m = 14

C - m = 5

D - m = -6

B - m = -5

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

ExplanationExplanation

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 = 3h’(3) = f(g(3)) - 6 = f(4) -6 = -1 - 6 = -7m = (y2 - y1) / (x2 - x1) = (-7 - 3) / (3 - 1) = -5