Exam 2: Functions and Graphs

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In exercise physiology, aerobic power P is defined in terms of maximum oxygen intake. For altitudes up to 1,800 meters, aerobic power is optimal-that is, 100%. Beyond 1,800 meters, P decreases linearly from the maximum of 100% to a value near 40% at 5,000 meters. Estimate aerobic power at the altitude of 2,400 meters.

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E

Find a formula that expresses the fact that P(x, y) is a distance 2 away from the origin.

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E

Find the domain of ff . f(x)=86x2+18x−108f ( x ) = \frac { 8 } { 6 x ^ { 2 } + 18 x - 108 }

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C

Simplify the difference quotient f(x+h)−f(x)h, if h≠0\frac { f ( x + h ) - f ( x ) } { h } \text {, if } h \neq 0 for f(x)=x2+4f ( x ) = x ^ { 2 } + 4

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Find (f∘g)(x)( f \circ g ) ( x ) . f(x)=x2−4x, and g(x)=x+9f ( x ) = x ^ { 2 } - 4 x \text {, and } g ( x ) = \sqrt { x + 9 }

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Find (f+g)(3)( f + g ) ( 3 ) f(x)=8x+2f ( x ) = 8 x + 2 , and g(x)=x2g ( x ) = x ^ { 2 }

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Find a composite function form for y if: y=4+x2+3y = 4 + \sqrt { x ^ { 2 } + 3 }

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The volume of a conical pile of sand is increasing at a rate of 576π576 \pi ft 3 /min, and the height of the pile always equals the radius r of the base. Express r as a function of time t (in minutes), assuming that r = 0 when t = 0.

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Find the distance d (A, B) between the points A (1, - 1) and B (8, 1).

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Sketch the graph of the function. f(x)=3x2−6x−5f ( x ) = 3 x ^ { 2 } - 6 x - 5

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Several values of two functions T and S are listed in the following tables: t 0 1 2 3 T(t) 2 3 1 0 X 0 1 2 3 S(x) 1 0 3 2 Find: (S∘S)(3)( S \circ S ) ( 3 )

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Find an equation of the line shown in the figure. Find an equation of the line shown in the figure.

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Find the domain of fgf g . f(x)=3xx−5, and g(x)=xx+1f ( x ) = \frac { 3 x } { x - 5 } , \text { and } g ( x ) = \frac { x } { x + 1 }

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Let y = f (x) be a function with domain D = [ -10, -9 ] and range R = [ -10, -7 ]. Find the domain D and range R for the function y=∣f(x)∣y = | f ( x ) |

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Find the domain of  f/g \text { f/g } . f(x)=x+7, and g(x)=x+7f ( x ) = \sqrt { x + 7 } , \text { and } g ( x ) = \sqrt { x + 7 }

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If the point P (3, 1) is on the graph of a function f, find the corresponding point on the graph of the function y=−9f(x−2)+1y = - 9 f ( x - 2 ) + 1

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Find (g∘f)(x)( g \circ f ) ( x ) f(x)=6x−7f ( x ) = 6 x - 7 , g(x)=4x2−x+2g ( x ) = 4 x ^ { 2 } - x + 2

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Find the distance d (A, B) between the points A (6, 6) and B (6, - 4).

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Find (f/g)(5)( f / g ) ( 5 ) . f(x)=−x2f ( x ) = - x ^ { 2 } , and g(x)=6x−2g ( x ) = 6 x - 2

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An object is projected vertically upward with an initial velocity of v0\mathcal { v } _ { 0 } ft/sec, and its distance s(t)s ( t ) in feet above the ground after tt seconds is given by the formula s(t)=−16t2+v0ts ( t ) = - 16 t ^ { 2 } + v _ { 0 } t . If the object hits the ground after 1515 seconds, find its initial velocity v0\mathcal { v } _ { 0 } in ft/sec.

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