Let f be a function defined on the closed interval [0, 1] such that f(0) = f(1) = 0 and f(x) = f(1-x) for all x in [0, 1]. Show that there exist at least four distinct values of c in [0, 1/2] such that f''(c) = 0.
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Answer:
Theorem: Let f:[0,1]→[0,1]. Assume f is continuous. Then there exists c∈[0,1] such that f(c)=c. • For any function f, let g(x)=f(x)−x. If the theorem ...
Answer:
(y
2
−y
1
)=m(x
2
−x
1
)
(y
2
+2)=2(6−1)=y
2
=8
So f(6) will be 8 or greater than 8.