- #1
John O' Meara
- 330
- 0
Suppose that a function f is differentiable at x0 and define g(x)=f(mx+b), where m and b are constants. Prove that if x1 is a point at which mx1+b=x0, then g(x) is differentiable at x1 and g'(x1)=mf'(x0).
Definition: A function f is said to be "differentiable at x0" if the limit
[tex] f'(x_{0})= lim_{h \rightarrow 0} \frac{f(x_{0}+h)-f(x_{0})}{h}[/tex]
exists. If f is differentiable at each point of the open interval(a,b), then we say that f is "differentiable on (a,b)", and similarily for open intervals of the form [tex] (a,+\infty),(-\infty,b)and (-\infty,+\infty)[/tex]. In the last case we say f is "differentiable everywhere". I think this is a simple proof, but since I am doing this no my own with a book that doesn't do too many example proofs, I am lost. g(x1) = f(mx1+b) = f(x0), which I think proves the limit of g(x1) exists but g'(x1)=? Thanks for the help.
Definition: A function f is said to be "differentiable at x0" if the limit
[tex] f'(x_{0})= lim_{h \rightarrow 0} \frac{f(x_{0}+h)-f(x_{0})}{h}[/tex]
exists. If f is differentiable at each point of the open interval(a,b), then we say that f is "differentiable on (a,b)", and similarily for open intervals of the form [tex] (a,+\infty),(-\infty,b)and (-\infty,+\infty)[/tex]. In the last case we say f is "differentiable everywhere". I think this is a simple proof, but since I am doing this no my own with a book that doesn't do too many example proofs, I am lost. g(x1) = f(mx1+b) = f(x0), which I think proves the limit of g(x1) exists but g'(x1)=? Thanks for the help.