a. use the mean-value theorem
f(2)=5 and f(5)=2
=> (f(5)-f(2))/(5-2)=(2-5)/(5-2)=-1
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
g'(x)=f'(x)*f'(f(x))
So
g'(2)=f'(2)*f'(f(2))=f'(2)*f'(5)
g'(5)=f'(5)*f'(f(5))=f'(5)*f'(2)
Therefore g'(2)=g'(5)
c. If f"(x)=0 ∀x, then
f(x)=mx+b, m,b∈ℝ.
f'(x)=m
g'(x)=m*m(f(x))=m² (m is not dependent on x)
=> g"(x)=0
d. h(x)=f(x)-x
=> h(2)=f(2)-2=5-2=3
and
=> h(5)=f(5)-5=2-5=-3
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.
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.
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