Calculus : Please show work as organized as possible. (I get easily confused with numbers). Please put a box on the answer to separate from the rest and put what question number it is. Thank you so much!

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1. [-/10 Points] DETAILS SESSCALC2 1.1.019. If f(x) = 3x2 - x + 5, find the following. f(1 ) = f(-1) = f(a ) = f(-a) = f(a + 1) = 2f(a ) = f(2a) = f(a 2 ) = [f(a) 12 = f( a + h ) = Need Help? Read It Watch It20. [-/5 Points] SESSCALCZ 1.3.537.XP. ASK YOUR TEACHER E The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion 5 = 2 sin at + 3 cos nt, where t is measured in seconds. (Round your answers to two decimal places.) (a) Find the average velocity during each time period. 0) [1,2] cm/s (ii) [1, 1.1] cm/s (iii) [1, 1.01] cm/s (iv) [1, 1.001] cm/s (b) Estimate the instantaneous velocity of the particle when t = 1. cm/s Need Help? _' 21. [-/6 Points] DETAILS SESSCALC2 1.3.AE.005. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER EXAMPLE 4 Investigate the following limit. lim sin _ x - 0 X SOLUTION Again the function f(x) = sin(5x/x) is undefined at 0. Evaluating the function for some small values of x, we get f(1) = sin 5x = (2 = sin 107 = 0 = sin 157 = = sin IT = 0 f(0.1) = sin 50m = 0 f(0.01) = sin 500m = Similarly, f(0.001) = f(0.0001) = 0. On the basis of this information we might be tempted to guess that lim sin 57 = x - 0 but this time our guess is wrong. Note that although f(1) = sin 5nn = for any integer n, it is also true that f(x) = 1 for infinitely many values of x that approach 0. You can see this from the graph of f given in the figure. The compressed lines near the y-axis indicate that the values of f(x) oscillate between 1 and -1 infinitely often as x approaches 0. (Use a graphing device to graph f(x) and zoom in toward the origin several times. What do you observe?) Since the values of f(x) do not approach a fixed number as x approaches 0, x - 0 lim sin 2 does not exist. Need Help? Read It22. [-/6 Points] DETAILS SESSCALC2 1.4.001. Given that lim f(x) = 4 lim g(x) = -2 lim h(x) = 0, x - 1 x-1 find the limits, if they exist. (If an answer does not exist, enter DNE.) (a) lim [f(x) + 3g(x)] x- 1 (b) lim [g(x)13 x - 1 (c) lim vf(x) X- 1 (d) lim 4f ( x ) x-1 g(x) (e) lim 9 ( x) x- 1 h(x) (f ) lim 9( x)h(x) x- 1 f(x) Need Help? Read It Watch It23. [-/6 Points] DETAILS SESSCALC2 1.4.002. The graphs of f and g are given. Use them to evaluate each limit, if it exists. (If an answer does not exist, enter DNE.) y = f(x) y = g(x) 0 (a) lim [f (x) + g(x)] x - 2 (b) lim [f(x) + g(x)] x - 1 (c) lim [f(x)g(x)] x - 0 (d) lim f ( x ) x--1 g(x) (e) lim [x3f (x ) ] (f) lim v 3+ f (x) X-1 Need Help? Read It Watch It25. [-/1 Points] DETAILS SESSCALC2 1.4.011. Evaluate the limit, if it exists. (If an answer does not exist, enter DNE.) lim X2 - 9x + 18 X - 3 X - 3 Need Help? Read It 26. [-/1 Points] DETAILS SESSCALC2 1.4.013. Evaluate the limit, if it exists. (If an answer does not exist, enter DNE.) lim x2 - 8x + 12 x - 4 X - 4 Need Help? Read It 27. [-/1 Points] DETAILS SESSCALC2 1.4.015. Evaluate the limit, if it exists. (If an answer does not exist, enter DNE.) t2 - 9 lim t - -3 2+2 + 7t+ 3 Need Help? Read It Watch It 28. [-/1 Points] DETAILS SESSCALC2 1.4.017. Evaluate the limit, if it exists. (If an answer does not exist, enter DNE.) lim (-9 + h)2 - 81 h - 0 h Need Help? Read ItEvaluate the limit, if it exists. (If an answer does not exist, enter DNE.) Iim \\/4+h2 hao h Need Help? Evaluate the limit, if it exists. (If an answer does not exist, enter ONE.) 3 _ lim X_'9 9x x2 Need Help? _' _' Evaluate the limit, if it exists. (If an answer does not exist, enter DNE.) lim (1 27 ) t>O t t +t Need Help? _' _' Evaluate the limit, if it exists. (If an answer does not exist, enter ONE.) 1 1 . + llm 8 x X"8 8+x Need Help? _' _' 33. [-/1 Points] DETAILS SESSCALC2 1.4.026. MY NOTES ASK YOUR TEACHER PRA Evaluate the limit, if it exists. (If an answer does not exist, enter DNE.) lim Vx2 + 25 - 13 x- -12 x + 12 Need Help? Read It 34. [4/6 Points] DETAILS PREVIOUS ANSWERS SESSCALC2 1.4.031. MY NOTES ASK YOUR TEACHER PRAC Use the Squeeze Theorem to show that lim (x2 cos 9xx) = 0. Illustrate by graphing the functions f(x) = -x2, g(x) = x2 cos 9xx, and h(x) = x2 on the same screen. Let f(x) = -x2, g(x) = x2 cos 9xx, and h(x) = x2. Then -1 0 S cos 9xx s 1 - f(x)