Math Help Forum - View Single Post

Mathstud28 · #2 May 11th, 2008, 10:32 PM

Quote:

Originally Posted by chousta

hi i need help with these asap

please...

1.
If

, then

is continuous at

Choice

0

everywhere

2.
The equation

implicitly defines

as a function of

. Find

.

3.

is discontinuous at

____ because _________________
(More than one selection may be correct. Keep in mind that if a function tends to infinity as

, the limit does not exist at

.)
Choice

because

does not exist.

because

does not exist.

because

does not exist.

because

does not exist.

because

Number of available correct choices: 2

4.

If

,
then we have

(Select the correct statement(s).)

Choice
f is both continuous and differentiable at x = 0
f '(x) exists for all x except x = 0 f is not continuous at x = 0
f '(x) is continuous for all x
f '(x) does not exist anywhere
f '(x) exists for all x but is not continuous for all x
f is continuous at x = 0 but not differentiable at x = 0
Number of available correct choices: 2

5.

Above is the graph of a function

. Sketch the graph of

?
6.

If

,
then at

0,

is??

7.

Find

if

.

8.

. Select the intervals on which the function is continuous.
Choice

Number of available correct choices: 3

Is this a test?

I will do ONE until you say it is not

8. $y=\tan\bigg(\frac{1}{(x+1)^4}\bigg)$

So $arctan(y)=\frac{1}{(x+1)^4}$

Differentiating we get $\frac{y'}{1+y^2}=\frac{-4}{(x+1)^5}$

Seeing that $y=\tan\bigg(\frac{1}{(x+1)^4}\bigg)$

wee see that $y'=\frac{-4\bigg[1+\tan^2\bigg(\frac{1}{(x+1)^4}\bigg)\bigg]}{(x+1)^5}$