Suppose that \(f \colon [0,2] \to \mathbb{R}\) is continuous on \([0,2]\) , differentiable on \((0,2)\), and such that \(f(0) = f(2) = 0\), \(f(c) = 1\) for some \(c \in (0,2)\). Prove that there exists \(x \in (0,2)\) such that \(|f'(x)| >1.\)
Proof. We will consider three cases. First, suppose \(c<1\). Then, by the mean value theorem, there exists \(x\in (0,c)\) such that \(f'(x)(c-0)=f(c)-f(0)\) so \(f'(x)=\frac{f(c)}{c}=\frac{1}{c}>1\) since \(c<1\). Similarly, if \(c>1\) then by the mean value theorem there exits \(y\in (c,2)\) such that \[|f'(y)|=\left\lvert\frac{f(2)-f(c)}{2-c}\right\rvert=\left\lvert \frac{-f(c)}{2-c}\right\rvert=\left\lvert\frac{-1}{2-c}\right\rvert>1\] since \(1<c<2\).
Now, suppose \(c=1\). If there exists \(x\in (0,1)\) such that \(x<f(x)\) then by the mean value theorem on the interval \((0,x)\) there exists \(s\in (0,x)\) such that \(f'(s)=\frac{f(x)}{x}>1\) since \(f(x)>x\). Likewise, if there exists \(x\in (0,1)\) such that \(x>f(x)\) then the mean value theorem on \((x,1)\) gives a point \(t\in (x,1)\) such that \(\left\lvert f'(t)\right\rvert=\left\lvert \frac{f(1)-f(x)}{1-x}\right\rvert=\left\lvert\frac{1-f(x)}{1-x}\right\rvert>1\) since \(x>f(x)\). So, on \((0,1)\), if the proposition does not hold then \(f(x)=x\). Similarly, if there exists \(x\in (1,2)\) such that \(f(x)>2-x\) then the mean value theorem yields a point \(u\in (x,2)\) such that \(|f'(u)|=\left\lvert \frac{f(2)-f(x)}{2-x}\right\rvert=\left\lvert \frac{-f(x)}{2-x}\right\rvert>1\) since \(f(x)>2-x\). If there exists \(y\in (1,2)\) such that \(f(y)<2-y\) then again by the mean value theorem there exists \(v\in (1,y)\) such that \(|f'(v)|=\left\lvert\frac{f(y)-f(1)}{y-1}\right\rvert=\left\lvert\frac{f(y)-1}{y-1}\right\rvert>1\) since \(f(y)<2-y\) so \(|f(y)-1|>|y-1|\). So, on \((1,2)\) if the proposition does not hold then \(f(x)=2-x\). However, notice that since \(f(x)\) is differentiable at \(x=1\) we cannot have \(f(x)=x\) on \((0,1)\) and \(f(x)=2-x\) on \((1,2)\). ◻
Solution
1. Setup: \(f\) continuous on \([0,2]\), differentiable on \((0,2)\), \(f(0) = f(2) = 0\), and \(f(c) = 1\) for some \(c \in (0,2)\).(The card already contains a complete case-analysis proof; this solution restates it in structured form.)
given.
2. If \(c < 1\): there is \(x \in (0, c)\) with \(f'(x) = \frac{f(c) - f(0)}{c - 0} = \frac{1}{c} > 1\).
Mean Value Theorem on \([0, c]\); since \(0 < c < 1\), \(1/c > 1\).
3. If \(c > 1\): there is \(y \in (c, 2)\) with \(|f'(y)| = \left|\frac{f(2) - f(c)}{2 - c}\right| = \left|\frac{-1}{2-c}\right| = \frac{1}{2-c} > 1\).
Mean Value Theorem on \([c, 2]\); since \(0 < 2 - c < 1\), \(1/(2-c) > 1\).
4. It remains to handle \(c = 1\).
4.1. If for some \(x \in (0,1)\), \(f(x) > x\): then \(f'(s) = \frac{f(x) - f(0)}{x - 0} = \frac{f(x)}{x} > 1\) for some \(s \in (0, x)\).
Mean Value Theorem on \([0, x]\); \(f(x) > x\) gives \(f(x)/x > 1\).
4.2. If for some \(x \in (0,1)\), \(f(x) < x\): then \(|f'(t)| = \left|\frac{f(1) - f(x)}{1 - x}\right| = \left|\frac{1 - f(x)}{1 - x}\right| > 1\) for some \(t \in (x, 1)\).
Mean Value Theorem on \([x, 1]\) (note \(f(1) = f(c) = 1\)); \(f(x) < x\) implies \(1 - f(x) > 1 - x > 0\), so the ratio is \(> 1\).
4.3. Hence, unless we are already done, \(f(x) = x\) for all \(x \in (0,1)\).
1 and 2 exhaust \(f(x) > x\) and \(f(x) < x\).
4.4. Similarly on \((1, 2)\): unless done, \(f(x) = 2 - x\) for all \(x \in (1,2)\).
if \(f(x) > 2 - x\) for some \(x \in (1,2)\), MVT on \([x,2]\) gives \(|f'(u)| = \left|\frac{f(2) - f(x)}{2-x}\right| = \frac{f(x)}{2-x} > 1\); if \(f(x) < 2 - x\), MVT on \([1, x]\) gives \(|f'(v)| = \left|\frac{f(x) - f(1)}{x - 1}\right| = \frac{|f(x) - 1|}{x-1} > 1\) since \(|f(x) - 1| > |x - 1|\).
4.5. \(f(x) = x\) on \((0,1)\) and \(f(x) = 2 - x\) on \((1,2)\) contradict differentiability at \(1\).
the left derivative of \(x\) at \(1\) is \(1\), the right derivative of \(2 - x\) at \(1\) is \(-1\); they differ, so \(f\) is not differentiable at \(1\).
5. Q.E.D.: in every case some \(x \in (0,2)\) has \(|f'(x)| > 1\).
2, 3, and 4 (cases \(c < 1\), \(c > 1\), \(c = 1\)) are exhaustive.