Posted by Leanna on Monday, February 7, 2011 at 10:53pm.
Let f be a twice-differentiable function such that f(2)=5 and f(5)=2. Let g be the function given by g(x)= f(f(x)).
(a) Explain why there must be a value c for 2 < c < 5 such that f'(c) = -1.
(b) Show that g' (2) = g' (5). Use this result to explain why there must be a value k for 2 < k < 5 such that g"(k)= 0.
(c) Show that if f"(x) = 0 for all x, then the graph of g does not have a point of inflection.
(d) Let h(x) = f(x) - x. Explain why there must be a value r for 2 < r < 5 such that h(r) = 0.
I know you have to use the intermediate value theorem and mean value theorem but don't know how.
- Calculus - MathMate, Monday, February 7, 2011 at 11:26pm
a. use the mean-value theorem
f(2)=5 and f(5)=2
It is a straight application of the mean-value theorem to state that there exists 2<c<5 such that f'(c)=(f(5)-f(2))/(5-2).
b. Use the chain rule to differentiate g(x) to get
c. If f"(x)=0 ∀x, then
g'(x)=m*m(f(x))=m² (m is not dependent on x)
Since f(x) is continuous and differentiable in (2,5), h(x) is also continuous and differentiable. Since h(r)=0 lies between h(2) and h(5) so the intermediate value theorem applies directly to state that 2<r<5 exists such that h(r)=0.
Post if you need more details.
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