<p>k, i have no idea what to do with these problems. I am not trying to leech answers, I hit a dead end. Here goes:</p>
<p>lim<br>
h->0<br>
(|x+h|-|x|)/h when x=3<br>
a) -1
b) 0
c) 1
d) 3
e) non existant</p>
<p>I got as far as removing the absolute value sign from the second x but i dont know what to do with the |x+h|.</p>
<p>Are you supposed to be solving this algebraically or applying calculus concepts to it? This equation actually turns out to be a derivitive question. Remember that the formula for the derivitive is
f'(x) = lim (h-->0) (f(x+h) - f(x)) / h.
That means that this question is simply asking for the derivitive of abs(x) when x=3.</p>
<p>which would be 0</p>
<p>Yes, tanman is correct.</p>
<p>Although there is a MUCH easier way to do it, there is a reason you do it out at the beginning.</p>
<p>o stupid me, i forgot to convert it to standard derivative form. Thanks much.</p>
<p>Why isn't the answer 1?</p>
<p>If f(x+h)=x+h, doesn't that mean that f(x)=x? So isn't the derivative 1? My cal I is a little rusty, so I dunno.</p>
<p>jpps1, I agree with you.</p>
<p>yea the derivative is 1</p>
<p>It's 1. Even if you suck at using the equation, try to visualize the graph, which has a slope of -1 from (-inf.,0), a point of discontinuity at x=0, and a slope of 1 from (0, inf.).</p>
<p>The x=3 is almost irrelevant, as long as it is greater than 0, the derivative will be one.</p>
<p>ok here's another question:</p>
<p>suppose that f is a continous function defined for all real numbers x and f(-5)=3 and f(-1)= -2. If f(x)=0 for one and only one value of x, then which of the following could be x?
a) -7
b) -2
c) 0
d) 1
e) 2</p>
<p>f(x) is not a straight line as far as I have figured so I'm guessing its a cubic type of function where the local max is below the x axis.</p>
<p>B. Try graphing the points and it should be pretty obvious. Since f is continuous, the function must pass over the x-axis at least once between x=-5 and x=-1. Since f only intercepts the x-axis once, only the answer between -1 and -5 is possible.</p>
<p>Whether it's cubic doesn't really matter at all. It could very well be just a straight line.</p>
<p>Just to give you a hint, look up the intermediate value theorum</p>
<p>Edit: never mind.. answer already posted above</p>
<p>o tru. I made it a straight line and kept getting a value of 2.6. thats why I was confused. Thx.</p>
<p>Oh, sorry. I wasn't paying attention. Yeah, it couldn't be a straight line, but it still didn't matter much what type of function it was.</p>