I think you could do it like this:

Let f(x) be the odd function we’re gonna work with. So, f(-x) = -f(x), By the definition of the derivative:

f'(x) = ##lim_(h->0)(f(x+h) – f(x))/h##.

We want to prove that f'(-x) = f'(x). Now, f'(-x) is:

##f'(-x) = lim_(h->0)(f(-x+h)-f(-x))/h## .

Since f(x) is and odd function, f(-x) = -f(x), therefore, we can rewrite the above expression as:

##f'(-x) = lim_(h->0)((-f(x-h)) + f(x))/h## = ##lim_(h->0)[-(f(x-h)-f(x))/h]##.

But we can rewrite the above expression as:

##f'(-x) = lim_(h->0)(f(x-h)-f(x))/-h##.

To make things clearer, let t = -h. Therefore:

##f'(-x) = lim_(t->0)(f(x+t)-f(x))/t##, which is just f'(x). That is: f'(-x) = f'(x). And there we have it: if f(x) is an odd function, then f'(x) is an even function.

Hope it helps you!

 

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How do you tell if your data from a table (in a lab) is directly or inversely proportional?

Solution: It’s very simple whenever in the lab you prepare a table you have two values. Suppose xand y are these two values. Now you are changing the value of x and correspondingly takingthe…

 

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How do you tailor an argument to SOAP in a business letter? Select all that appl;y.

A. through content 

B. through style

C. by figuring out what you would want to hear from someone trying to convince you

D. none of the above

 

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##x = 5/8##

You used Logarithm of a product (based on the symbol you used which is addition).

##log2 + log(4x-1) = 3####log 2(4x-1) = 3####log (8x-2) = 3####8x-2 = 3####8x = 3+2####8x = 5##

(then divide them with 8 so that we can remove the value of x which is not applicable), so..##8x-:8 = 5-:8##

##x = 5/8##

 

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Answer is 11, You use the L shaped box (read below)

So first step is to create that L shaped box and write 3 outside of it.

The first number is the coefficient of the greatest power, or 4, since it’s with ##x^2##. Go in order, the next number should be the coefficient of x, and the next after that is ##x^0=1##.

Pull the first number down without doing anything to it, then multiply that by the number outside the Lso ##4 * 3 = 12##take that answer and put it back in under the second coefficient and add them together-9 + 12 = 3

then take that answer, multiply by the number outside the box##3 * 3 = 9##and put that under the third coefficient and add them together2+9=11

Plug x=3 into the equation to check.

 

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