Exam 1 Feb 22, 2007 SHOW ALL WORK Math 25 Calculus I Either circle your answers or place on answer line. Find the following derivatives (you do not need to simplify):

[14] 1.) \frac{d}{dx}[\frac{x^{2}+3\sqrt{x}+x}{2x^{4}−5}] Answer 1.)

[14] 2.) \frac{d}{dx}[2xe^{x} + 3\sqrt{x^{5}} − \frac{1}{x}] Answer 2.)

3.) Calculate the appropriate limits in order to find the equations of all vertical and horizontal asymptotes for f(x) = \frac{\sqrt{x^{2}+1}}{2(x−3)}. Show ALL steps.

[12] horizontal asymptotes)

[10] vertical asymptotes)

[10] 4a.) Find the derivative of f(x) = 2x + 3 by using the definition of derivative.

f′(x) =

[3] 4b.) Find the equation of the tangent line to the curve f(x) = 2x + 3 when x = 1.

[10] 5.) Express the given quantity as a single logarithm.:

a ln(x) + b ln(y) − c ln(z) − d ln(1) =

[7] 6.) Sketch the graph of a function with the following properties:

lim_{x→2^{+}}f(x) = +∞, lim_{x→+∞}f(x) = 0, lim_{x→−∞}f(x) = 5 f′(−3) = 1, f′(0) = 0, f′(1) = −4 7.) If a ball is thrown vertically upward with a velocity of 16 ft/sec, then its height (in feet) is given by s(t) = 16t − 16t^{2}.

[7] 7a.) What is the maximum height reached by the ball?

[3] 7b.) Find a point (t_{0}, s(t_{0})) at which the slope of the tangent line to the curve s(t) = 16t − 16t^{2} is equal to 0: \underline{(t_{0}, s(t_{0})) =}

[10] Choose either problem 8 or 9. You may do both problems for up to 4 points extra credit.

8.) Let f : R → R, f(x) = (x − 3)^{2}.

8a.) Is f 1:1? . If f is not 1:1, prove it.

b.) Is f onto? . If f is not onto, prove it.

9a.) State the Intermediate Value Theorem.

9b.) Use the Intermediate Value Theorem to show that \sqrt{x} − 3 = 0 has a root between 4 and 16.