For each function, find (a)ƒ(x+h), (b)ƒ(x+h)-ƒ(x), and (c)[ƒ(x+h)...

Question

For each function, find (a)ƒ(x+h), (b)ƒ(x+h)-ƒ(x), and (c)[ƒ(x+h)-ƒ(x)]/h.See Example 4. ƒ(x)=1/x^2

Accepted Answer

Hey, everyone here we are asked to solve for F of X plus H F of X plus H minus F of X and the quantity of F of X plus H minus F of X divided by H for the following function F of X is equal to two divided by X squared. Now, for this problem, we have four different answer choice options, A B C and D and each of the answer choices have three different solutions. One for each of our three expressions. So we can start by finding our first expression, which is F of X plus H. Well, to find F of X plus H, all we have to do is substitute in X plus H four X from our original function. So instead of having two divided by X squared, we will now have two divided by equality of X plus H squared. So there isn't any simplifications that we need to do with this expression. So this expression is our solution for F of X plus H. Now we can move on to finding our second expression, which is F of X plus H minus F of X. Now to find this expression, all we need to do is substitute in our expression that we found for F of X plus H as well as the given expression for F of X. So doing this, we again can begin with F of X plus H which is two divided by the quantity of X plus H squared. And now we have minus F of X. So we just have minus two divided by X squared. Now, we just need to simplify this expression by combining the fractions. But before we can combine the fractions, we need to get common denominators for both of the fractions. So our first fraction needs to be multiplied by the denominator of our second fraction. So two divided by the quantity of X plus H squared gets multiplied by X squared divided by X squared. And now our second fraction gets multiplied by the denominator of the first fraction. So two divided by X squared gets multiplied by the quantity of X plus H squared divided by the quantity of X plus H squared. So now we just need to simplify this expression by first multiplying the numerator and the denominator is together. So we have two X squared divided by X squared times the quantity of X plus H squared. And then we have minus two times the quantity of X plus H squared, all divided by X squared times the quantity of X plus H squared. And now our next step here is to just factor out our X plus H squared in the second numerator as well as distribute in the negative two. So doing this, our expression is two X squared divided by X squared times the quantity of X plus H squared minus or plus negative four, negative two X squared minus four X H minus two H squared, all divided by X squared times E quantity of X plus H squared. And now we can just simplify for further by combining our new operators. So we have two X squared minus two X squared minus four X H minus two H squared all divided by X squared time, the quantity of X plus H squared. So now we need to continue to simplify this expression by combining the like terms in the numerator. Here, we can see that both of the two X squares cancel out. So our simplified answer for F of X plus H minus F of X becomes H times the quantity of negative four X minus two H all divided by X squared times the quantity of X plus H squared. So now that we have found our simplified form of our second expression, we can move on to finding our third expression where we have the quantity of F of X plus H minus F of X all divided by H. So we just need to repeat a similar process as previously where we substitute in the known expression. And then here we need to divide it by H. So we just found F of X plus H minus F of X. So we first need to just substitute in this expression. So we have H times the quantity of negative four X minus two H all divided by X squared times the quantity of X plus H squared. And this is all divided by H Well, we know that when we divide by a value, we can also multiply by the reciprocal. So rewriting our expression, we have H times the quantity of negative four X minus two H all divided by H times X squared times the quantity of X plus H squared. And now we can make some simplifications. Here, we see that we have an H in the numerator and in the denominator. So both of these terms cancel out. And from this, we see that our final simplified form of the quantity of F of X plus H minus F of X all divided by H simplifies to the quantity of negative four X minus two H all divided by X squared times E quantity of X plus H squared. So here we have found our three simplified expressions for the three required expressions. And with these three solutions, we are left here with answer choice B where F of X plus H is two divided by the quantity of X plus H squared. Then F of X plus H minus F of X is the quantity of H times the quantity of negative four X minus two, two H all divided by X squared times the quantity of X plus H squared. And finally, the quantity of F of X plus H minus F of X all divided by H gives us the quantity of negative four X minus two H all divided by X squared times the quantity of X plus H squared. Thanks for watching. I hope you found this video helpful and I'll see you again next time.

For each function, find (a)ƒ(x+h), (b)ƒ(x+h)-ƒ(x), and (c)[ƒ(x+h)-ƒ(x)]/h.See Example 4. ƒ(x)=1/x^2

Key Concepts

Function Evaluation

Difference Quotient

Limit Process