Problem

Show that a cubic function (a third-degree polyno…

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Problem 84 Hard Difficulty

(a) Show that $ e^x \geqslant 1 + x $ for $ x \geqslant 0 $.
(b) Deduce that $ e^x \geqslant 1 + \frac{1}{2} x^2 $ for $ x \geqslant 0 $.
(c) Use mathematical indcution to prove that for $ x \geqslant 0 $ and any positive integer $ n $,
$$ e^x \geqslant 1 + x + \frac{x^2}{2!} + \dots + \frac{x^n}{n!} $$

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Chapter 4

Applications of Differentiation

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Video Transcript

Okay, so we're being asked to show that even X is greater than one. Playtex were all expected. Indio. So we're going to assume that ever vex is equal to E T X minus one plus x. So this completely normal? You can do this. You can assume any function. So let's take the derivative a crime of X will that could just e to the X minus one. And we know that ae to the access always greater than one for all Ex greater than zero. And you know this because off the graph of your backs. So if that's true, we know that prime has toe also be greater than zero on the same domain on X rated zero. So if that's true, if it's increasing, we know that Affect is also positive. It's not necessarily positive, but it's greater than F zero because F zero is the minimum point. So we know that every vexed or any number greater than X, which is all that's written. Zero. This is true. So in June bug and half of zero, well, that just f of X. And that's just going to be greater than zero, because after a zero zero Oh, after Jiro is Yeah, on. So then we can If we know that's true, then we could not to go ahead and plug in half of excellent give. Each of the X minus one plus X is greater than zero. And now we can move, determines over to the other side and get one plus X. And it is proven for be well being has to prove e to the X is greater than one plus one half y squared for all X rated. You know, we're going to do really similar a proof style. We're gonna let FX hope that is well, that affects ik o e t x minus one plus X plus one x squared. And then if you take the derivative of that brown of X equals you did the X minus one plus X. Now this looks really familiar and in fact, this it is that is very familiar to our f effect. And we proven that effort back is also great, isn't zero, so we can actually imply that each of the exes also greater than one plus X. So then that implies that FX again is greater than F zero, which is just zero and so we can plug in our function already looking to e to the X minus one plus x plus one half x squared and this is greater than equal to zero. And then you can rewrite this by bringing it to the Sergeant Got heated. Jax. It's greater than one plus x plus one half x squared. And that's the proof. And for the last one, we're being told to use mathematical induction to prove that for X rated Indio and in any positive integer and so induction. If you're not familiar, what it is, you're basically proving that if it worked for us a any, but to say that this proof work for So for this case, we have easy X ray, the one plus X plus X squared over to factory and then on and on to infinity or whatever, an integer actually and ran factorial. So if you think about it, we have just proved it for any holes zero on and it was doing factory or in equals one factorial, which was just this case right here. And so what induction means is that we're trying to prove it for the end plus one or the end. So if we know it for the ones we know that it worked. For one. The question is, what does the work for any clothes to work for three days or four days? Five on and on and on. So basically and plus one and essentially all mathematical induction is It's a very clever idea. I think this may be a little bit harder to digest, but just think about the ideas. So we're going to do something again. Very similar with every Rex equals a Juliet minus one plus X plus one half, Um, this is one of factorial act square and then plus, and then it's going to be, um we're going to let it equal one over and plus one factorial because we're looking for not just e end. Um, we're not just doing it for one and reporting it for all end. So but by adding and plus one reprove ing that this work for any end and so we're going also placed Ex city and plus one. Now we take the derivatives of the derivative. It's going to be eating the eggs, and it will be minus blinding plus Ah, gonna be one plus X plus one over to factorial n squared. Plus it is going to be one over in factorial times X to the end. We already know that for the regular case exit E end over in factory. We know that that's Oh, no, I'm sorry. We know that f prime is greater than zero. No, So we know after prime is greater than zero from previous proof, so we can actually go ahead and plug in that f of X. It's great, isn't it? Walk West Africa zero. But in this case would be zero. And so we can go ahead and even clogging our function. So e to the X minus one plus x plus one over to factorial. And that's X squared plus and then one over in plus one factorial accident plus one masqueraded and zero. And then we moved over all that. Yeah, all that term. So one plus x plus one over to factorial three one over and plus one factorial and an exit E and plus one. So that's the proof

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Problem

Show that a cubic function (a third-degree polyno…

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Video Transcript

James Stewart

Calculus: Early Transcendentals

Anna Marie Vagnozzi

Kayleah Tsai

Caleb Elmore

Michael Jacobsen

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Anna Marie Vagnozzi

Kayleah Tsai

Caleb Elmore

Michael Jacobsen

James StewartCalculus: Early Transcendentals

Anna Marie Vagnozzi

Kayleah Tsai

Caleb Elmore

Michael Jacobsen

James Stewart

Calculus: Early Transcendentals