Continuity question

[imwishing73]

If h(x)={3x^2-2, x<-1} {f(x) , x>-1}, then h(x) will be a continuous function at x=-1 when f(x)=
a. 2^(x-1)
b. (x^3)-1
c. sin(x+1)
d. cos(x+1)
e. x+1

This appeared on an old test, and the answer is cos(x+1). Now during midterm review, I'm struggling to figure out how I did this. In graphing the answer, it does not appear continuous to me, and none of the other answers look continuous to me, so graphing my options is not a viable method. I wouldn't even think that a trigonometric function would be used here. How exactly does one go about solving such a problem, or any similar problem?

 

[srmichael]

If h(x)={3x^2-2, x<-1} {f(x) , x>-1}, then h(x) will be a continuous function at x=-1 when f(x)=
a. 2^(x-1)
b. (x^3)-1
c. sin(x+1)
d. cos(x+1)
e. x+1

This appeared on an old test, and the answer is cos(x+1). Now during midterm review, I'm struggling to figure out how I did this. In graphing the answer, it does not appear continuous to me, and none of the other answers look continuous to me, so graphing my options is not a viable method. I wouldn't even think that a trigonometric function would be used here. How exactly does one go about solving such a problem, or any similar problem?

This is a piecewise graphing problem, yet you don't have to graph it in order to answer it. In order for the graph to be continuous, h(x) must equal f(x) at x = -1, i.e. h(-1) = f(-1). Thus one can draw the entire graph without lifting their pencil (i.e. there is no discontinuity)

Knowing this, you then calculate h(-1) = 3(-1)² - 2 = 3-2 = 1. The only one of the five options where f(-1) = 1 is (d) since cos(-1 + 1) = cos(0) = 1

 

[imwishing73]

Wow I feel dumb. I need to get out of this brainlock before the actual midterm...thanks!