(adsbygoogle = window.adsbygoogle || []).push({}); A few calculus questions....

Hey all. I have a few questions involving calculus. Any help would be much appreciated.

1) [tex] f(x)= \left\{\begin{array}{cc}x^2,&\mbox{if} x \leq 0\\3x+1, &\mbox{if} 0<x<4\\12-x^2,&\mbox{if} x>4\end{array}\right [/tex]

Find the following limits,if they exist:

i) [tex]\lim_{x\rightarrow 2} 3x+1 [/tex]

[tex] = 3(2) +1 [/tex]

[tex] \lim_{x\rightarrow 2} 7 [/tex]

ii) [tex] f(0) [/tex]

replacing every equation with 0, i got [tex] x^2= 0, 3x+1= 1, 12-x^2= 12 [/tex] the answer i got was that the limit does not exist. Would this be true in this case?

2) Find the slope of the tangent line to the curve [tex] y=x^3-x [/tex] at the point (-2,-6) using:

i) [tex]\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h} [/tex]

ii) [tex]\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a} [/tex]

i) Using the first formula I got an answer of 11 so technically, using the second formula should give me the same the answer but I can't get it and I have to show work. Here is what I have so far:

[tex] \lim_{x\rightarrow -2} \frac {x^3-x-((-2)^3-(-2))}{x+2} [/tex]

[tex] \lim_{x\rightarrow -2} \frac {x^3-x+6}{x+2} [/tex]

Not applying L'Hospital's Rule and solving this by factoring, I need to get a [tex] (x+2) [/tex] on the top so the top and bottm [tex] (x+2) [/tex] equal one. And then I can put in a -2 for x since you're not going to be dividing by zero. I hope I make sense and could anyone tell me if I'm doing something wrong? Thanks.

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# Homework Help: A few calculus questions

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