Use the 4 step process to find f(prime)(x) then fine f(prime)(1), f(prime)(2), f(prime) 3

Eq: f(x) = 2 + (3/x)

f(prime)(x) = ?

Can someone help please?

3 posts
• Page **1** of **1**

Use the 4 step process to find f(prime)(x) then fine f(prime)(1), f(prime)(2), f(prime) 3

Eq: f(x) = 2 + (3/x)

f(prime)(x) = ?

Can someone help please?

Eq: f(x) = 2 + (3/x)

f(prime)(x) = ?

Can someone help please?

- jpd75
**Posts:**4**Joined:**Thu Jan 14, 2010 7:43 pm

What is the "four-step process" that you are supposed to be using?

Please show all your work so far, when you reply.

Please show all your work so far, when you reply.

- nona.m.nona
**Posts:**249**Joined:**Sun Dec 14, 2008 11:07 pm

I don't know what is meant by the 'four step process', perhaps the four steps I used to get the answer is what they want. Here's how I would do the problem...

f'(x)=2+(3/x) becomes f(x)=2+3^-1, so then f'(x)= -3^-2 or f'(x)=-3/x^2

or else the 'four step process' might mean to find it the long way, although this requires way more than 4 steps:

{lim as h->0} [f(x+h)-f(x)]/h

-fill in f(x+h) and f(x) to get

{lim as h->0} [(2+3/x+h)-(2+3/x)]/h

-find the common denominator for each parenthetic term in the numerator to get

{lim as h->0} [((2x+2h+3)/(x+h))-((2x+3)/x)]/h

-find common denominator for the entire numerator to get

{lim as h->0} [(x(2x+2h+3)/x(x+h))-((x+h)(2x+3)/x(x+h))]/h

-expand the numerator of the numerator to get

{lim as h->0} [(2x^2+2hx+3x-2x^2-2xh-3x-3h)/x(x+h))]/h

-Simplify the numerator of the numerator to get

{lim as h->0} [-3h/x(x+h)]/h

-copy dot flip (multiply numerator by 1/h) to get

{lim as h->0} -3h/hx(x+h)

-cancell out the h's to get

{lim as h->0} -3/x(x+h)

-set h equal to zero to get

-3/x^2

since that was WAY MORE than 4 steps, you're probably safer using the first example.

f'(x)=2+(3/x) becomes f(x)=2+3^-1, so then f'(x)= -3^-2 or f'(x)=-3/x^2

or else the 'four step process' might mean to find it the long way, although this requires way more than 4 steps:

{lim as h->0} [f(x+h)-f(x)]/h

-fill in f(x+h) and f(x) to get

{lim as h->0} [(2+3/x+h)-(2+3/x)]/h

-find the common denominator for each parenthetic term in the numerator to get

{lim as h->0} [((2x+2h+3)/(x+h))-((2x+3)/x)]/h

-find common denominator for the entire numerator to get

{lim as h->0} [(x(2x+2h+3)/x(x+h))-((x+h)(2x+3)/x(x+h))]/h

-expand the numerator of the numerator to get

{lim as h->0} [(2x^2+2hx+3x-2x^2-2xh-3x-3h)/x(x+h))]/h

-Simplify the numerator of the numerator to get

{lim as h->0} [-3h/x(x+h)]/h

-copy dot flip (multiply numerator by 1/h) to get

{lim as h->0} -3h/hx(x+h)

-cancell out the h's to get

{lim as h->0} -3/x(x+h)

-set h equal to zero to get

-3/x^2

since that was WAY MORE than 4 steps, you're probably safer using the first example.

- sdbielz
**Posts:**6**Joined:**Sun Jul 19, 2009 4:53 pm

3 posts
• Page **1** of **1**