Recent content by Crazy_Diamond

Ah I forgot about that. I end up with a and 1/2x. And substituting 2 will give me 1 which is the right answer! Thank you so much for your help. I have just question though. Finding the derivative gives me f'(x) so substituting x = 2 is the solution because doing so gives me the limit as x...

But how can I solve for a if c is in the reduced function? Since by letting ##f(2^-) = (f(2^+)## will give me ##4a = 1 - 3c##, I still cannot solve for a due to the unknown c.

I end up with ##c = \frac{1 - 4a}{3}## What should I differentiate? I differentiated this to end up with: ##c = \frac{\frac{1 - 4a + h}{3} - \frac{1 - 4a}{3}}{h}## ##c = 1/3## but plugging that into the former gives me a = 0

Oh I see. So to make it factorable, let b = 4a and that way it becomes: ##\frac{a(x^2 - 4)}{x - 2}## ##a(x+2)## and now ##a(x + 2) = 1 - 3c## How can this be solved?

I don't understand. How can there be a "hole" if ##\frac {1} {4} x^2 - 3c## has a value at 2? Doesn't the question imply that it's a continuous function?

It's dicontinuous? I'm aware that "2" is actually not the value but 1.999... but I don't know how to express that in a mathematical way. I'm not really sure actually. For to cancel out x - 2 it cannot be equal to 2 which it already is not according to the question right? But I'm not sure what...

True sorry about that. I'm not sure what happens when ##\frac{4a - b}{2 - 2}## now as I don't know how to factor out (x-2) from the original ##\frac{4a - b}{x - 2}##

In that case I would receive for left side: f(2-) = 4a - b / 2 - 2 = 4a - b, x =/= 2 and for right side: f(2+) = 1 - 3c So since the limits on both sides of 2 must be equivalent, I will get: 1 - 3c = 4a - b But now how can I solve for each variable?

Ah sorry about that. Thank you for moving my thread.

Homework Statement The problem is: f(x) = {ax^2 - b / (x - 2), x < 2 1/4(x^2) - 3c, x >= 2 If f is differentiable at x=2, what are the values of a, b, and c? In case it is hard to see, I have provided an image 2. Homework Equations I know that the limit definition of a...

I spoke with my guidance counselor and he told me that, because I'm far into my program, I won't be able to switch into another discipline (systems, mechatronics, and electronics) due to them being too full and resources are low. At this point I'm stuck in the mechanical engineering program...

I've heard a lot about engineering physics. I don't have that option at Waterloo but I've been advised by most peers and teachers to avoid it as its job prospects aren't too favorable. Though they are all the same people that advised my to take mechanical as it's safe but now that I'm...

This is a great idea that I am considering. Ideally though I'm not sure if I can take a double major in both mechanical engineering and physics. Additionally my mechanical engineering program doesn't offer many electives but instead chooses a list of electives that I can choose from...

Hello everyone, I'm currently an undergraduate student who just completed his first year in mechanical engineering. The reason I'm making this thread is because I'm not satisfied with learning classical physics. I'd also like to learn and apply modern physics as well. Because of my...