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Explain the difference between finding $f(0)$ and finding the input $x$ for which $f(x)=0$...

Question

Explain the difference between finding $f(0)$ and finding the input $x$ for which $f(x)=0$

Explain the difference between finding $f(0)$ and finding the input $x$ for which $f(x)=0$



Answers

Explain the difference between finding $f(0)$ and finding the input $x$ for which $f(x)=0$

We have a question in which we need to get a mind whether actors directors are not. So we have effects which is equal to access squired sign one by X. When X is not equal to zero and zero of annex equal to zero. Okay, so we need to find um we need to differentiate first and we need to find left and innovative. So if differentiation if derivative exists so left and derivatives should be able to write and innovative. So left and derivative will be Limit. X approaches to zero from left side, affects minus, affects minus f zero by x minus zero. 40 by X zero. No, if you convert it into which it will become Limit accidents to zero F zero minus set plus f zero by zero minus at minus zero, 0- at minus F zero x 0- set minus zero as we are approaching X from the left side. Okay, so this is Limit at approaches to zero. F off minus set -F of zero, divided by minus that. So let us plug in the value in the function at approaches to zero F minus set is minus that whole square Sign -1 by H zero because 0 is zero. My mindset. So this will become limited accidents to zero at a square Sign -1 by H by -H. So I wanted to and actually get canceled out. We'll be getting and one more thing sign of any negative number sign -1 battled mind sign it. So this limit as approaches to 0- at sign But one by H Divide by- at -1. Yeah. Okay so finally we'll be getting at sign one batch now if that approaches to zero. Uh So this will become zero into Sign one by H. So we know that range of signage. A range of sine function is always -1-1. So any value of one batch will give us between any number between -1-1. So this is zero. So left unlimited zero similarly right different derivatives right And elevating will be limit at approaches to zero from right side, F X -F0 By X zero. So this will be limit. Sorry here actually approaching 20 limit Such approaches to zero, plus H minus f zero but express edge. Uh It should not be a plus at it should be zero plus at Yeah. Okay. Zero blacks. Okay in the denominator it should be zero plus eight minus zero. Triple A. Search zero. So this will be a limit. That approach is 20 F h minus F zero by H. Now plugging in the function Limited approaches to zero and 2 square sign one by it and F 00. So let it be much. Now again this will tend to zero and this will tend Any value between my 1-10. So we can see that left and derivative equal to write and debating hands. The limit exists. Thank you him limited. Very very big. Sorry, thank you.

He It's clear. So when you raid here, So we're gonna take the derivative. Do you ever d x for X square over one plus eat Vex. You were gonna ply the quotient rule, which is equal to D over DX X square one plus B to the X minus d over dx one plus b to the x Times X square all over one plus e to the x square We simplify and we get to x times one plus eat the x minus Eat the X Times X square all over one plus eat the X square for the second derivative. We're also gonna ply the quotient rule Let me get tea over d x for two x times one plus eats the X minus eats the x x square comes one plus Eat the X Square dynasty over d x one plus e to the x square times two x comes one plus heat the x minus Eat the X Times X square This is all over one plus. Eat the X square and square that again, and when we simplify we get is this equal to eat? Two X times X square minus Eat the X X Square minus four e to the two x times X minus four e to the x Times acts plus four e to the X Plus two e two the two x plus two all over one plus eat the X cubed.

It's clear. So when you read here, so we have f of X is equal to X over X square minus one. We're gonna use the quotient role. We'll get X square minus one D acts d over DX minus x times D over D X X square minus one well over X square minus one square just becomes equal to X square minus one minus two X square over X square minus one square, which is equal to negative X square minus one over X square minus one square. We're gonna find the second derivative by using the Kocian rule, which is X square minus one square D over DX your negative X square minus one minus negative X square minus one de over de x X square minus one square All over X square minus one square. You square that again, this becomes equal to that's square minus one times negative. Two acts less negative. Two X square minus two turns negative. Two acts well over X square minus one. Cute. You could simplify this to two x times X square plus three all over X square minus one. Cute

This question asked us to find the function f Such a f prime of X is X f of X minus acts and then given the fact that F zero is too now, first things first. We know we can right f prime of acts in terms of de roi de axe. It's easier to use why instead of F because then we can later on integrate our variables factor the right hand side to make it easier to divide. Now we can transfer all the Y terms over to the left hand side and all the ex terms over to the right hand side. Take the integral of both sides Integrate. Remember, we increased the extranet by one as you can see it from X to X squared and then we divide by the new exponents which is to ad see witches are constant integration. Remember that we're substituting x zero unwise to more solving for C so literally just substitute in our values To get rid of the variables and solve for C, we have zero equal see substitute sequel zero back in Now we know we're not done. Remember, the game plan is we need to get why equals? Which means we have to raise are based e We're raising the power to the base E on each equivalent side. So we have Y minus one is e to the X squared over to remember each. The Ellen is simply one. It just cancels. Now there's one last thing we have to do. Remember how the problem initially gave this in terms of F of X? Well, what this means is that we need to substitute why with FX, and then we will be flashed.


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