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Select the one function that is NOT continuous on the interval (2,9) The slope of the tangent line to y=f(x) at the point where x=5 can be computed using which one of the following limits? \fExplain why the function is discontinuous at the given number a. (Select all that apply.) f(x) = 1 X+2 a = -2 Of(-2) and lim f(x) exist, but are not equal. X->-2 V x->- 2+ lim f(x) and lim f(x) exist, but are not equal. X->-2- O f(-2) is undefined. lim f(x) does not exist. X--2 none of the above XExplain why the function is discontinuous at the given number a. (Select all that apply.) x2 - 2x f(x ) x2 - 4 if x * 2 a = 2 1 if x = 2 O lim f(x) and lim f(x) are finite, but are not equal. x- 2+ X- 2- f(2) is undefined. V X- 2 lim f(x) does not exist. f(2) is defined and lim f(x) is finite, but they are not equal. X-2 O none of the above XHow would you "remove the discontinuity" of f? In other words, how would you define f(4) in order to make f continuous at 4? f ( x ) = X - 2x - 8 X - 4 f(4) = 0\fy by the Intermediate Value Use the Intermediate Value Theorem to Show that there is a root of the given equation in the specied Interval. x4+x9=o, (1,2) f(x) = x4 + x 9 is _y on the closed interval [1, 2], F(1) = X , and 2) =l = 0 In the Interval (1, 2). Theorem. Thus, there Is a root v J of the equation X4 + x 9 J . So it is undefined when cos(x) = , and this happens when x = 1/, +3/, .... Thus F has discontinuities when x is an odd multiple of It and is continuous on the intervals between these values (see the figure).For the function f whose graph is given, state the limits specified below. Enter co or -co when applicable. (a) lim f(x) X - co (b) lim f(x) X - -00 (c) lim f(x ) x- 1 (d) lim f(x) X- 3 (e) the equations of the asymptotes (Enter your answers as a comma-separated list of equations.) vertical horizontalFind the limit. (Let a and b represent arbitrary real numbers. If an answer does not exist, enter DNE.) lim X ->00 x- + ax - V x2+ bx 0 X