Exam 3: Functions

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Find the domain. g(t)=1t2−7t+12g ( t ) = \frac { 1 } { t ^ { 2 } - 7 t + 12 }

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Let f (x) = x 2 . Find and simplify the indicated difference quotient. f(y+h)−f(y)h\frac { f ( y + h ) - f ( y ) } { h }

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Using the results in the table, graph the functions y = x 2 , y = x 2 - 2, and y = x 2 + 2 on the same set of axes. x X 2 X 2 - 2 X 2 + 2 0 0 -2 2 ±\pm 1 1 -1 3 ±\pm 2 4 2 6 ±\pm 3 9 7 11  Using the results in the table, graph the functions y = x<sup> 2 </sup>, y = x<sup> 2 </sup> - 2, and y = x<sup> 2 </sup> + 2 on the same set of axes. x X<sup> 2 </sup> X<sup> 2 </sup> - 2 X<sup> 2 </sup> + 2 0 0 -2 2  \pm  1 1 -1 3  \pm  2 4 2 6  \pm  3 9 7 11

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Specify the domain and range of the given function. (The axes are marked in one-unit intervals.) Specify the domain and range of the given function. (The axes are marked in one-unit intervals.)

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Assume that the accompanying viewing rectangle shows the essential features of the graph of f(x) = -x 3 + 4x+ 2. Use a graphing utility to estimate to the nearest hundredth the coordinates of the turning points. Assume that the accompanying viewing rectangle shows the essential features of the graph of f(x) = -x<sup> 3 </sup>+ 4x+ 2. Use a graphing utility to estimate to the nearest hundredth the coordinates of the turning points.   f(x) = -x<sup> 3 </sup>+ 4x+ 2 [- 4, 4, 2] by [- 12, 12, 6] f(x) = -x 3 + 4x+ 2 [- 4, 4, 2] by [- 12, 12, 6]

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Suppose that in a certain biology lab experiment, the number of bacteria is related to the temperature T of the environment by the function N(T)=−2T2+240T−5,100(40≤t≤90)N ( T ) = - 2 T ^ { 2 } + 240 T - 5,100 ( 40 \leq t \leq 90 ) Here, N(T) represents the number of bacteria present when the temperature is T degrees Fahrenheit. Also, suppose that t hr after the experiment begins, the temperature is given by T(t)=10t+40(0≤t≤5)T ( t ) = 10 t + 40 ( 0 \leq t \leq 5 ) How many bacteria are present when t = 5 hr?

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Assume that the domain of ff and f−1f ^ { - 1 } is (−∞,∞)( - \infty , \infty ) . Solve for t. f(6−2t6+2t)=12;f−1(12)=−3f \left( \frac { 6 - 2 t } { 6 + 2 t } \right) = 12 ; f ^ { - 1 } ( 12 ) = - 3 .

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Find the domain. g(t)=t2−9t+203g ( t ) = \sqrt [ 3 ] { t ^ { 2 } - 9 t + 20 }

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Find the domain. H(t)=5t−47−2t3H ( t ) = \sqrt [ 3 ] { \frac { 5 t - 4 } { 7 - 2 t } }

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Let T (x) = 4x 2 - 3x. Find (and simplify) the expression. T (x + h) - T (x - h)

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Set up a table and graph the function y=1/x.y = 1 / x .

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Sketch the graph. y={21−x2 if −1≤x<12x if x≥1y = \left\{ \begin{array} { l l } 2 \sqrt { 1 - x ^ { 2 } } & \text { if } - 1 \leq x < 1 \\\frac { 2 } { x } & \text { if } x \geq 1\end{array} \right.

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Given that f and g are inverse functions. If f(7) = 12, what is g(12)?

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Use a graphing utility to graph the function and then apply the horizontal line test to see whether the function is one-to-one. y = x 7 + 5x A) True B) False

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Find the domain. h(x)=x2−49h ( x ) = \sqrt { x ^ { 2 } - 49 }

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Determine the domain and the range. F(x)=x3+6x3−2F ( x ) = \frac { x ^ { 3 } + 6 } { x ^ { 3 } - 2 }

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A spherical weather balloon is being inflated in such a way that the radius is given by r=g(t)=12t+2r = g ( t ) = \frac { 1 } { 2 } t + 2 Assume that r is in meters and t is in seconds, with t = 0 corresponding to the time that inflation begins. If the volume of a sphere of radius r is given by V(r)=43πr3V ( r ) = \frac { 4 } { 3 } \pi r ^ { 3 } find the time at which the volume of the balloon is 36πm336 \pi \mathrm { m } ^ { 3 }

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Let g(t)=4t+6g ( t ) = \frac { 4 } { t } + 6 . Find g - 1 (t).

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Set up a table and graph the function y=x2−2y = x ^ { 2 } - 2 .

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Find the domain. g(t)=t2−5t+6g ( t ) = \sqrt { t ^ { 2 } - 5 t + 6 }

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