I am in need of serious help. I am not sure how to do numbers 14, 18, 24, and 33

Question

I am in need of serious help. I am not sure how to do numbers 14, 18, 24, and 33. I would really appreciate an explanation because my test is next week.

7, a Is f defined at 27 (L.ook at the definition of fa 35. lin d is continuous on and why? b. Is f continuous at-2 S. At what values of x is fcontinuous? 9. What value should be assigned to 112) to make the function continuous atx27 37. 10. To what new value should J(1) be changed to remove the y = g(x) tinuity? 38. Co 39 Applying the Continuity Test At which points do the functions in Exercises 11 and 12 fail toe tinuous? At which points, if any, are the discontinuities r Not removable? Give reasons for your answers. I1. Exercise 1, Section 2.4 12. Exercise 2, Section 24 fail to be At what points are the functions in Exercises 13-30 continuous 13. y=x-2 15.y= k(x) 14, y= + 4 14. y= 16, y= x2-3x-10 18. y = 20), y = cosx 22. y= tan- 24, y = 1+sin-x 26. y=Vir-1 28, y = (2-x)1/ - 3x 22+4 (x + 2)2 x + 3 15. 4x + 3 17. y=1x-1| + sin x cos r x + 2 19, y= -x- 21. y=csc 2x r tan x 23, y= 25, y=V2x + 3 27, y = (2-1)1/3 r2 1 1 + sinx x-3 5, 29. g(x) =

Explanation / Answer

14)y=[1/(x+2)2] +4

given function is discontinous if denominator equals 0 because function is undefined

(x+2)2=0

x+2=0

x=-2

limx->-2-[1/(x+2)2] +4 =[1/(-2+2)2] +4=

limx->-2+[1/(x+2)2] +4 =[1/(-2+2)2] +4=

function has non removable dicontinuity at x =-2

actually function has infinate discontinuity at x =-2

18)y =(1/(|x|+1)) -(x2/2)

is dicontinous if denominator =0

|x|+1=0

but it is not true, it never happens as |x| always greater than zero

function is always continous

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