Posted by Leanna on Monday, February 7, 2011 at 10:53pm.
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.
Calculus (Continuity and Differentiability) - Okay. So I am given a graph of a ...
Math--Calculus - I'm having a tough time figuring out this problem... S(x...
mean value theorem - Show that the function f(x)=1-|x|, [-1,1] does not satisfy ...
calculus ap - If g is a differentiable function such that g(x) is less than 0 ...
Calculus - Suppose f(x) is a differentiable function with f(-1)=2 and f(2)=-1. ...
calculus - The function f is twice differentiable, and the graph of f has no ...
calculus - Let f be a differentiable function such that f(3) = 2 and f'(3...
Calculus - Decide if the following function f(x) is differentiable at x=0. Try ...
mat 116 algebra - function and function notation using the variable x and ...
Calculus - flickr.(dotcom)/photos/77304025@N07/6782789880/ Create a function ...
For Further Reading