calc help

Test Preparation
1

<p>i wanna know if im doing this hmwk right. plz tell me what u get.
piece function so its all on the same graph
f(x) = x^2-1, -1(less than or equal to)x<0;
f(x)=2x, 0(less than or equal to)x<1;
f(x)=1, x=1;
f(x)=-2x+4, 1<x<2;
f(x)=0, 2<x<3</p>

<p>1 a. does f(-1) exist?
b. does lim as x approaches -1 frm the right of f(x) exist?
c. does lim as x approaches -1 from the right of f(x) = f(-1)?
d. is f continuous at x=-1?</p>

<li>a. does f(1) exist?
b. does lim as x approaches 1 of f(x) exist?
c. does lim as x approaches 1 of f(x) = f(1)?
d. is f continuous at x=1?</li>
</ol>

<p>3 a. is f defined as x = 2?
b. is f continuous at x = 1?</p>

<p>im not really sure if im doing this right so thank you so much if you will tell me what you, i’ve allready put down answers but im not sure if they are correct.</p>

2

<p>1a. f(-1) exists
b. yes
c. yes
d. yes (if the equation continues to the left, is that specified?)</p>

<p>2a. yes
b. yes
c. no
d. no</p>

<p>3a. no
b. no (same as 2d?); it is not continuous at x=2 either because we don't know f(2)</p>

3

<p>i got all the same cept 3b. why is it yes?</p>

4

<p>Wait, if 3b is the same question as 2d, why would 2d be no but 3b be yes? And why did they ask it twice?</p>

5

<p>wait i mean 2b. and 3b is suppose to be is f continuous at x = 2</p>

6

<p>grr umm im looking over a test we got back in calc and i want to make sure what the correct answer is. sorry i need so much help cause i learn nothing in that class</p>

<ol>
<li><p>find lim as x approaches 3 of the sqaure rt of (9-x^2)</p></li>
<li><p>find lim as x approaches 2 of (x-2)/(absolute value of x-2)</p></li>
<li><p>find lim as x approaches 0 of ((sqaure rt of (x+9))-3)/x</p></li>
<li><p>find lim as x appraches 1 of 5/((x-1)^2)</p></li>
<li><p>find lim as x approaches 1 of (2-(5/((x-1)^2))</p></li>
</ol>

7

<p>2b The limit as x approaches one exists because the limit is 2 from both the right and the left. The fact that f(1)=1 doesn't matter, because it still appears as if the line goes to 2 from either side.</p>

<ol>
<li>plug in 3, sqrt(9-9) equals 0</li>
<li>does not exist, because limit from left = -1, limit from right = +1</li>
<li>plug in 0, sqrt(0+9) - 3 = 3-3 = 0</li>
<li>inifinity (graph it)</li>
<li>still infinity, it just moves up 2 (infinity+2 :D )</li>
</ol>

8

<p>i got 9. wrong and i had put 0</p>

<p>and all these last questions frm last test was done multiple choice in the ap format without a calculator so how would u kno 14. without a graph?</p>

9

<p>You graph it by hand. It's a really simple graph: you know there is no value for two, which means the graph does not cross it. So either both sides are going to infinity, both are going to negative infinity, or they're going in opposite directions along the asymptote. So you pick -1, it's 5/9, 0 is 5, it's going up. Same for the other side, it's going up. Therefore, infinity.</p>

<ol>
<li>DNE. I forgot, 0 from the left doesn't exist because then the square root is negative.</li>
</ol>

10

<p>Plugging in 0 for (9) gives you 0/0, an undeterminate form. You need to use L'Hospital's rule, if you know it.</p>

<p>The answer is not necessarily DNE. For example</p>

<p>lim (x->3) (x^2-9)/(x-3) </p>

<p>gives you 0/0 when plugging in 3 into the expression, but it has a limit.</p>

11

<p>wait why can't there be a value for two? if u plug in 2 u get 5, or do u mean 1 is not a value because its the asy.?</p>

12

<p>we don't learn l'hosptial's rule till the end of the course.</p>

13

<p>When we take the limit of a function at a value, we could care less what the function actually is at that value. It could be 0,2,5, ininity, not existing, it doesn't matter. We only care about the values of the function very very close to the value, but never actually at it. That's why it doesn't matter what happens when you plug in 2.</p>