I need some calculus help....

Question

anonymous · Answer

What's your question?

irishboy123 · Answer

$$ f(x)=\dfrac{x^2}{x^2+81} = 1- 81(x^2+81)^{-1} $$
$f' = 81(2x)(x^2+81)^{-2}$

is that where you are because it looks like you are correct?

thecalchater · Answer

Find the value of derivative (if it exists) of the function f(x)7-|x| at the extremum point (0,7)

thecalchater · Answer

0 
Does not Exist
-8 
8

anonymous · Answer

what is f(x)?

anonymous · Answer

oh okay

anonymous · Answer

so it is f(x)= 7- abs{x} ?

irishboy123 · Answer

$(|x|)'$ is not defined

anonymous · Answer

the derivative is abs{x}/x

anonymous · Answer

sorry I forgot the minus

anonymous · Answer

it does exist

thecalchater · Answer

can you put the steps and answer and walk me through it

anonymous · Answer

abs{x}= x for x>=0 and -x for x=<0

thecalchater · Answer

so 0?

thecalchater · Answer

it only count as participation, but I want to understand it

amistre64 · Answer

what does it mean to 'be a limit'?

amistre64 · Answer

hmm, well good try at defining it.  i was leaning more towards ... a limit is defined if and only if all paths lead to the same thing as we approach it.

amistre64 · Answer

now for your case, a derivative is a limit ... the limit of a slope.  is the slope from the left and the right the same as we approach 0,7?

thecalchater · Answer

yes?

thecalchater · Answer

(will you be able to help me with a few more after this one?)

amistre64 · Answer

graphing it may be useful ..

|dw:1449791491297:dw|

thecalchater · Answer

so no...

amistre64 · Answer

at the point 0,7. the slope from the left is 1, the slope from the right is -1.

since the value is not the same for all directions that we can approach from .. we say that the limit is not defined .. does not exist.

thecalchater · Answer

ok thanks ill post the second one give me a minute

thecalchater · Answer

locate the absolute extrema of the function f(x)=cos(pix) on the closed interval [0,3/4]

amistre64 · Answer

what should we test for?

thecalchater · Answer

if it is continous on the interval?

amistre64 · Answer

cos is a continuous function ... 

what else, regarding min and max do you think would be useful?

thecalchater · Answer

trying to find the mix and max on that interval?

amistre64 · Answer

yes, using the derivative set equal to zero we can determine some min/max critical points ... and also testing the end points.

thecalchater · Answer

ok so?

amistre64 · Answer

so what is our derivative of cos(pi x)

thecalchater · Answer

-pisin(pix)

amistre64 · Answer

and when that is set to equal zero, what do we get for our solution set?

thecalchater · Answer

my calc says +-2n, pi +-2n

thecalchater · Answer

that is where i am stuck

amistre64 · Answer

any integer value of x will make sin(pix)=0

thecalchater · Answer

so is it no max and min f(0)=1?

amistre64 · Answer

|dw:1449792069863:dw|