Just square both sides
I have another test question i got wrong and didnt undertand soo im hoping someone can help me out.
Show that sqrt(1+x) <1+(1/2)x if x>0 or ( formally written in attachment)
I used the formula f'(c)= [f(b)-f(a)]/(b-a) and then just ended up with a mess at the end which is probably why i got it wrong. How would i go abouts starting this problem?
You'd need a variation on that.
If you prove a derivative is always positive,
then you are proving the function is always increasing.
Instead....
Locate the maxmum value of g(x)
If
then if g(k)<0 the inequality is proven.
Alternatively
Find the value of x=k giving minimum h(x),
then show h(k)>0
x=0.
2nd derivative test at x=0.....
this is 0.25 at x=0, indicating a minimum on the curve.
h(0)=0
This means that if x is positive, h(x)>0
So im looking at my original question with the square roots and inequality and i dont see how i can take those values and sub then in the mean value Theorem.
I basically wote it as [1+(1/2)x - sqrt(1+x)]/ [(1/2)-(1+x)^(1/2)]... then eaither i got that wrong or i messed something up in simplifing because i couldnt prove thats its >0