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Let f(x)=square root (x). If the rate of change of f at x=c is twice its rate of change at x=1, then c=
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Aren't I looking for the second derivative and just looking for f " (1)?
no. I understand that as \[f(x)=\sqrt{x}\implies f'(x)={1\over 2\sqrt{x}}\\ f'(x=c)=2\times f'(x=1)\\ {1\over2\sqrt{c}}=2\times{1\over 2\sqrt{1}} \]then solve for 'c'
rate of change = first derivative
no wonder I kept getting -1/4 when the key stated it is +1/4 thanks, I will redo the problem
thank you
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you are welcome.
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