show that f(x) = r^5 + r^3 has no local extrama

Question

anonymous · Answer

extrema*

anonymous · Answer

I think If I am able to show that function is strictly increasing then it should work. A strictly increasing function cuts x axis only one time, I mean only one intercept on x-axis.

hmm for x intercept y= 0 or f(x)=0
f(x) = 0 
$$x^5 = -x^3 $$ 
Now through inspection we can see that only for x=0 the equation satisfies

amistre64 · Answer

you mean show that it just smooths down

amistre64 · Answer

id say it has no anything since there is no x value to plug in

amistre64 · Answer

f(x) = r^5 + r^3

f'(x) = r' (5r^4 + 3r^2)

amistre64 · Answer

but thats assuming r is a function of x; otherwise

f'(x) = 0

anonymous · Answer

ah okay I see now I took f(x) = x^5 + x^3

amistre64 · Answer

:)  yep

anonymous · Answer

hmm thanks :)

anonymous · Answer

a local extreme must have second derivative equal to zero and first derivative change signs. here, second derivative is onyl zero for x=0, and there, first derivative is 5x^4+3x^2, which is positive for both x=1 and x=-1 so it is an inflection point and not an extreme.