Show that #f# has no extremas? (Probably using Fermat's Theorem)
#f# is 2 times differentiable in #RR# with #f''(x)=0# , #x# #in# #RR#
& #f'(x)+x^2=(5-x)f(x)#
Show that #f# has no extremas?
Tip: Use Fermat theorem
&
Show that
Tip: Use Fermat theorem
2 Answers
There is no such function.
Explanation:
If
(This is a consequence of the Mean Value Theorem.)
As such, it has derivative
So either
However no linear function satisfies the given equation
If
#f'(x)+x^2 = (5-x)f(x)# becomes:
#m+x^2 = -(x-5)(mx+b)# which requires
Setting corresponding coefficients equal we get
and
To see that there is no such function without implicitly using the Mean Value Theorem, please see below.
Explanation:
Suppose that for all
Observe that
Differentiate both sides on the equation, to get
Use the two facts assumed to get:
Expand:
Solve for
But then, with a bit of work, (or electronic help)
Which is not identically
So