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znimon
A function is given below. Determine the average rate of change of the function between x = -3 and x = -3 + h. f(t) = √-7t
I have \[\sqrt{-7h}/h\] but it says that is wrong and I can't figure out why
Well, the 'average' rate of change for some interval \([a,b]\) (non-calculus, please tell me if you need otherwise) would be: \[ \Delta f_{avg}=\frac{f(b)-f(a)}{b-a} \]Try using that.
All right, then that should be the case.
so you are telling me my answer is right?
Nope, sorry. Using the above, we find, for \(f(t)=\sqrt{-7t}\) \[ \frac{f(-3+h)-f(-3)}{-3+h+3}=\frac{\sqrt{21-7h}-\sqrt{21}}{h} \]If you need further simplification of the above, please tell me.
how did you get -7h out from under the root?
How does one? You can't, you'd have to multiply both the numerator and denominator by \(\sqrt{21-7h}+\sqrt{21}\), but then it would end up on the denominator.
This is only useful for evaluating the limit.
I have no idea what you mean by that
why would I multiply the numerator and denominator by that?
My 7 sub 2 is \[\sqrt{21-7h}\]
If you wish to remove the \(h\) from the radical, you'd have to do that, but, of course, then the top expression ends up in the denominator. So, the point is that one cannot remove the \(h\) from such.
f(-3+h) = \[\sqrt{-7(3+h)}\]
My equation does not simplify. And, yes, that last statement is correct. Keep in mind: \[ \sqrt{a+b}-\sqrt{a}=\sqrt{b}\\ \]Is *not* necessarily true (In fact, it is mainly true if b=0 or a=0).
The original problem looks like the square root goes over the "t"
Yes, and that's how I computed it. What do you feel is wrong with my expression?
I don't understand how there is no square root sign over the 7h in your third comment
Where is there not a square root sign?
the 7h that is in the numerator of your third comment
http://imgur.com/1fwvB This is what I have in my browser and what has been typed.
oh that is weird it doesn't look like that in my browser
so my first comment is correct then
Square root of (-7h) divided by h
No, it is not, as they are not equivalent statements.
yeah it is because square root of (x+y) is equal to square root of x plus square root of y right?
No, it does not. \[ \sqrt{a+b}\ne\sqrt{a}+\sqrt{b} \]Unless a or b is zero.
oh jesus I feel like an idiot. So then your third comment does not simplify any further in pre calc?
Nope. I don't think there is any need to, unless you're taking limits.
so last thing square root of x*y is equal to square root of x times the square root of y?
Never mind I just proved it.
Thanks again I'll have to look up a khan academy video on that
Yes, that statement is true. Since: \[ a^2=n\\ b^2=m \]So we say: \[ nm=a^2b^2=(ab)^2 \]And all right, sure thing.