Suppose that $ f $ and $ g $ are continuous on $ [a, b] $ and differentiable on $ (a, b) $. Suppose also that $ f(a) = g(a) $ and $ f'(x) \leqslant g'(x) $ for $ a < x < b $. Prove that $ f(b) < g(b) $. [Hint: Apply the Mean Value Theorem to the function $ h = f - g $.]
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suppose that after NGO continuous on the closed interval from A to B and different won't it be supposed that ever be a good idea? And if prime of axis lesson equilar Jeep I make for a lesser necklace and B and we're being asked to prove this statement I hear effort These last Ngobe. Okay, so the first thing we have to do in this problem is we're going to assume what the statement says. A jingles G. So we're just going to rewrite that age of equals aftereffects. Mine is direct. Press it g Okay, now, since we are already told that this that these two functions are both continuous and defensible, we can always we can apply the mean value to him right away because those are the conditions required for the bathroom. So then apply it to H We'LL get a prime C equals age of B minus Asia of a all over B minus. And then we can rewrite this because we have a chew of excess. Simply assume h of X is equal effort back money. Jeeva. We can actually rewrite the left side and the right side will unite a prime of see it minus G primacy. And since h of b, we can also rewrite as effort B minus oh, minority of B. And this is one group spline minus. And we could do the same thing with FAA minus g ave. And just remember, these are just substitution we're doing. These are coming from the initial assumption we made earlier. Okay, now, that way we have the statement. We were also told that effort is decoded gov So we can automatically since f one the same number minus the number of zero. We can just we can put a big zero right here, because that is the same thing we're told that. And then what I'm going to do here is I'm gonna bring this B minus a on this side. So I'm gonna rewrite this. We're going to bring this to another page, Have B minus a time. Primacy minus she Privacy equal f A b minor jia f b minor Geo. Be sorry about that. And now, also going back to the first page. We're also told that F prime of X is less than equal to G prime of X. And since g prime is greater and it is a negative. This left hand side is negative and we can also assume the same thing for the right hand side. This also negative. So when something is negative, we know that is less than zero. So we can rewrite the right hand side as an inequality sign. So this is going to be less than zero because that's what we just as what this tells us. And then we can bring this over. You can add your baby and bam. We have now proven that G A B is greater than effort be for all ex and this for all acts in this interval from a TV.