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Mathematics 25 Online
OpenStudy (anonymous):

how do you find f^1(x) of f(x)=1/2x+2?

OpenStudy (anonymous):

f^-1(x) you mean?

OpenStudy (anonymous):

a claculator and a formula

OpenStudy (anonymous):

no, the derivitive.

OpenStudy (anonymous):

I know how to find it, the fraction is just throwing me off.

OpenStudy (anonymous):

the derivative of a degree 1 function (line) is the slope, 1/2

OpenStudy (anonymous):

Do you mean that f(x) = 1/(2x) + 2? Then write it as (2x)^(-1) if the fraction throws you off. Then use the normal derivative rule where you subtract the power of x by 1 etc, to get f'(x) = -2x^(-2).

OpenStudy (anonymous):

it's the whole, f(x+h) plug it in to f(x+h) - f(x) all over h? that's how our teachers been teaching us..

OpenStudy (anonymous):

the difference quotient with h approaching 0?

OpenStudy (anonymous):

is it 1 over 2x or one-half x?

OpenStudy (anonymous):

is it \[f(x)=\frac{1}{2x+2}\]?

OpenStudy (anonymous):

or \[f(x)=\frac{1}{2}x+2\]

OpenStudy (anonymous):

second one will be quick first one will be a pain, so we hope it is the second. let us know

OpenStudy (anonymous):

\[f(x+h)-f(x)=\frac{1}{2}(x+h)+2-(\frac{1}{2}x+2)\] \[=\frac{1}{2}x+\frac{1}{2}h-\frac{1}{2}x-2=\frac{1}{2}h\] divide by h, get \[\frac{1}{2}\] because the derivative is the slope if it is the other one let me know and we can try that one too

OpenStudy (anonymous):

thank you so much for the help!

OpenStudy (anonymous):

i have more, i'll keep you updated, thank you!

OpenStudy (anonymous):

yw

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