I really could use help with these questions, they are killing me.

Weekly conceptual questions A. Which of the following correctly displays the graph of a line tangent to y = x at x = 2? a. b. C. d. ... 4+ 1. .. .. .2- A B. Which of the following methods will correctly help us find the value(s) of x at which the graph of the function y = g(x) has a horizontal tangent line(s)? a. Evaluate g' (0). b. Set g'(x) = 0 and solve for x. c. Evaluate g(0). d. Set g(x) = 0 and solve for x. e. None of the above methods will work.The graph of fOC) on the interval [2,3] is below. Use it to answer questions C-F. C. At which of the following values of x is f not continuous? Circle all that apply. a. 1 b. 0 c. 1 d. 2 e.fis continuous at all those values D. At which of the following values of x is f not differentiable? Circle all that apply. a. 1 b. 0 c. 1 d. 2 e.f is differentiable at all those values E. Based on the graph, estimate the value of f'(0). a. me) : 2 b. rm) : 0 0. me) : 1 d. f'(0) = 1 e. f'(0) does not exist. F. Based on the graph, estimate the value off'(2). a. {(2) = 2 b. f'(2) = 0 0. Hz) = 1 d. f'(2) : 1 e. f'(2) does not exist. 1. Let f(x) = _ _*+ 2x-8 and g(x) = x+4 x-2 a. Fill in the missing function values in the following table, write 'undefined' if the value does not exist. x 1.99 1.999 2.0 2.001 2.01 f(x) x 1.99 1.999 2.0 2.001 2.01 g(x) b. Answer each question either 'yes' or 'no' and then give a brief explanation of why that is the correct answer. i. Is it true that * + 2x-8 =x+4? x - 2 ii. Is it true that lim * + 2x -8 x-2 = lim(x +4) ? x - 2 iii. Is it true that lim * + 2x - 8 - = lim(x +4) ? x- 2 2. Evaluate lim x+4 x->4x2+3x -43. In questions 1 and 2 we looked at the limit of a function as the input approached a pre-chosen value. Now let's look at limits being used to calculate an instantaneous rate of change. We want to find out the rate at which P(x) = -2x +3x + 5 is changing when x = 2. Recall the Difference Quotient: DO = J (x+h)- f(x) h is the slope of the line that goes through the points (x, f(x)) and (xth, f(x+h)). This can be thought of as the rate of change of f(x) over the interval a. Calculate and then simplify the Difference Quotient for P(x) = -2x2 +3x+5. b. Evaluate the Difference Quotient from part (a) for x = 2 and h =1. DQ = We have just calculated the rate of change of P(x) over the interval c. Evaluate the Difference Quotient from part (a) for x = 2 and h = 0.1. DQ= We have just calculated the rate of change of P(x) over the intervald. Evaluate the Difference Quotient from part (a) for x = 2 and h = 0. l . DQ= We have just calculated the rate of change of P(x) over the interval e. Evaluate the Difference Quotient from part (a) for x = 2 and h = 0.001. DQ: We have just calculated the rate of change of P(x) over the interval f. Evaluate the Difference Quotient from part (a) for x = 2 and h = 0.001. DQ: We have just calculated the rate of change of P(x) over the interval g. Take the limit as h approaches 0 of the Difference Quotient. Limit of DQ as h approaches 0 is: Evaluate What you got at x = 2. DQ evaluated at x : 2 is: We have just calculated the rate of change of P(x) at the point 10- -3 -2 2 4 5 6 -6 -8 -10 5. The graph above is of y = g(x) Based on the graph, answer the following questions. a. lim g(x) = *-> - 2+ b. lim g(x) = * -> - 2- c. lim g(x) = d. g(4) = e. lim g(x) = *-> 4 f. lim g(x) = g. lim g(x) = -of x h. lim g(x) = i. lim g(x) = * -> 5 k. For what intervals of x is y = g(x) continuous?6) Given h(x) = 2x2 5x 6 . - fo+Ax)f(x) Calculate h'(x) using the denition of the derivative: f (x) 2 3) T Check your result by taking the derivative using the 'simple' rules from sec 1.6. Graph y1 = h(x) and below graph y2 = h' (x) ; comment on any correlations you notice between the 2 graphs. (for example: when h'(x) = O we see what attribute on the graph of h(x) ?)