Check work please ! How can I find where a function is decreasing in interval notation if I have no critical points f(x)=(7x-2)/(x+7) f'(x)=47/(x+7)^2 I found that its increasing at (-inf,-7)U(-7,inf) but I don't understand how to tell where it is decreasing
f'(x) = 47/(x+7)^2 can this be negative ever ?
assuming your math for first derivative is correct, what is the second? Do you use first or second deriv. to find dec. or inc.?
I used the first derivative
it could never be negative ?
that means, the slope can never be negative
wat does that tell about the function ? the function is always increasing right ?
may not be "strictly increasing", but it does tell u that the function is never decreasing
The function is alwas increasing, I put none as my answer but apparently its wrong since I thought the function only increased
You can check for critical pts on the first deriv using the second, maybe go back and check the original problem. Did you copy it correctly? Is the first deriv correct?
so let's redo the first deriv.
Can you tell me the quotient rule?
Okay I used quotient rule f(x)=(7x-2)/(x+7) f'(x)=(x+7)(7)-(7x-2)(1)/ (x+7)^2 and I would end up with actually 51/(x+7)^2
bottom * derivative top - top *derivative bottom, all over bottom squared ?
yep
Okay (x+7)*(7) - (7x-2)*(1)
So, you were right we get \(g'(x)=\frac{51}{(x+7)^2}\)
now, we know that assuming x is \(\underline{\color{red}{real}}\), this can never be negative.
BUT: We do have a singularity
Namely x=-7 As you pointed out above with your note of where it is increasing. @dnova21, is this for like a lon cappa or online system?
online system
so we decided there is no where where this is decreasing right?
When you type it in, try doing an empty set
Oh okay let me check
It worked thank you so much
np, I know how crappy those systems can be
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