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if f(1)=3 and f'(x)>(equal) 2 for 1<(eq)x<(eq)5, how small can f(5) possibly be?
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\[f(1)=3\] start at (1,3) you know the slope is at least 2 so just think of a line with m = 2. that is the smallest m can be. equation for that line would be \[y-3=2(x-1)\] \[y=2(x-1)+3\] so smallest y can be if x = 5 (i.e. f(5)) is \[2(5-1)+3=8+3=11\]
or think this. from 1 to 5 is 4 units. the slope is at least 2 so you have gone at least 8 units. you start at 3, you are now at least 11
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