Practicing Success
Let f : [0, 1] → [0, 1] be a continuous function. Then, |
f(x) = x for at least one x ∈ (0, 1) f(x) will be differentiable in [0, 1] f(x) + x = 0 for at least one x such that 0 ≤ x ≤ 1 none of these |
f(x) = x for at least one x ∈ (0, 1) |
Since, $f:[0,1] \rightarrow[0,1]$ is a continuous function. Therefore, Range $(f) \subset[0,1]$ and f(x) attains every value between f(0) and f(1) Let $g(x)=f(x)-x$ for all $x \in[0,1]$. Clearly, g(x) is also continuous on [0, 1] Also, $g(0)=f(0)-0=f(0)>0$ and, $g(1)=f(1)-1<0$ [∵ f(1) < 1] Thus, g(x) is continuous on [0, 1] such that g(0) g(1) < 0. Therefore, there exists a point $c \in(0,1)$ such that Therefore, there exists a point $c \in(0,1)$ such that $g(c)=0 \Rightarrow f(c)=c$ Hence, f(x) = x for at least one $x \in(0,1)$. |