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if f(x)=13-8x+root(2x^2) and f'(r)=4 find r?
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\[f(x) = 13 - 8x + \sqrt{2x^2}\] do you know how to find f(r)?
no, that's what is confusing me
hint: change all the x with r
So like this? \[f(x)=13-8\times4+\sqrt{2\times4 ^{2}}\]
f(r)*
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what did you do?
i said replace x with r...not 4
but since f(r)=4 I thought it was that
no...
just replace x with r
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\[13−8×r+\sqrt{2×r^{2}}\] ?
right
simplify that
\[13-8r+r \sqrt{2}\]?
yes.
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btw...are you sure the question is EXACTLY \[f(x) = 13 - 8x + \sqrt{2x^2}\]
yes, it's exactly that
you're sure... so \[f(r) = 13 - 8r + r\sqrt 2\] take the derivative of this
The answer is supposed to be: \[r=3\sqrt{2}\] but the derivative gives me: \[f'(r)=-7\]?
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