Another way to approximate derivatives is to use the centered ...

Question

Another way to approximate derivatives is to use the centered difference quotient: f' (a) ≈ f(a+h) - f(a- h) / 2h. Again, consider f(x) = √x.

c. Explain why it is not necessary to use negative values of h in the table of part (b).

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Given the function K of X is equal to E to the power of X, approximate the derivative at X equals 0, using the centered difference quotient. Why can you omit negative values of H from your table? Awesome. So it appears for this particular problem, we're trying to approximate the derivative of this function for KFX at X equals 0, using the centered difference quotient. That's our first answer that we're trying to solve for, and then our second answer is we're trying to explain why we can omit the negative values of H from our table. Awesome. So with that in mind, knowing now that we're signing for two separate answers, our first step that we need to take is we need to define our known variables and the target variable that we're trying to solve for. So our given function, which is known. is going to be K of X is equal to the power of X, and our point of interest is going to be X equals 0, and our target variable is we're trying to figure out the approximate derivative of K 0. Awesome. And we need to note and recall once again, we need to recall and use the centered difference quotient formula, which in this case, we will be able to write our centered difference quotient formula considering K of X. We need to write K. X is approximately equal to K X plus H minus K X minus H. All divided by 2 H. OK. Awesome. So once again, that's our centered difference quotient formula. And now our next step is we need to use our centered difference potion formula. So our first step for that is we need to or for using our formula is we need to substitute the value of X equals 0 into our formula, meaning we need to take K of 0 is approximately equal to. K of 0 plus H. And we need to subtract K of 0 minus H. All divided by 2 H. Fantastic. And our next step now is we need to simplify the function evaluations, meaning we need to write that K of 0 + H is going to be E to the power of H. And we also need to write that K of 0 minus H is going to be equal to E to the power of negative H. So when we substitute in these values back into our formula, we will find that K of 0 is approximately equal to E H minus E the power of negative H. All divided by 2 H. And now our next step is we need to explain why the negative values of H can be omitted. This is our second and final answer that we're ultimately trying to solve for. So we solve for our first answer. And now, in order to explain why we can omit the negative values of H, we need to recall and note that the centered difference quotient formula is symmetric with respect to H. So therefore when we use both positive and negative values of H, it will yield the same result due to the symmetry. So meaning E power of H minus E the power of negative H divided by 2H will be the same for H and negative H. So without a doubt, our final two answers will be our first answer is K 0 is approximately equal to eh minus E the power of negative H, all divided by 2 H. And our second answer, which is for why we can omit the negative values of H from our table. is due to the fact that positive and negative values of H will yield the same results due to symmetry, and that is officially our final answer for this problem. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Another way to approximate derivatives is to use the centered difference quotient: f' (a) ≈ f(a+h) - f(a- h) / 2h. Again, consider f(x) = √x.
c. Explain why it is not necessary to use negative values of h in the table of part (b).

Key Concepts

Centered Difference Quotient

Symmetry in Function Values

Behavior of the Function Near a Point

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