Let f be a differentiable function such that f(0) = -5 and f'(x) is less than or equal to 3 for all x. Of the following, which is not a possible value for f(2)? A) -10 B) -5 C) 0 D) 1 E) 2
Let's suppose \(f'(x)=3\) for all \(x\) (the maximum given). By the mean value theorem, over some interval \((0,k)\), with \(k>0\), there is some \(c\) such that \[f'(c)=3=\frac{f(k)-f(0)}{k-0}=\frac{f(k)+5}{k}~~\implies~~3k-5=f(k)\] Now if \(k=2\), we would get \(f(k)=1\). What does this suggest?
I'm not sure but I think it means that f(2) is less than or equal to 1?
Right! We choose the maximum value for \(f'(x)\) to guarantee an upper bound for the value of \(f(k)\), in this case \(f(2)\). We know that \(f(2)\le1\), which means \(f(2)\) cannot possibly be \(2\).
Oh, okay. Thanks!
You're welcome!
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