OpenStudy (anonymous):

f(x)= (1/2x)(ln(x^20)), (-1,0)…..find the equation of the tangent line. y=

6 years ago
OpenStudy (anonymous):

\[m _{t}=(f'(x)=0)\]

6 years ago
OpenStudy (anonymous):

???

6 years ago
OpenStudy (anonymous):

what class are you in

6 years ago
OpenStudy (anonymous):

calculus

6 years ago
OpenStudy (anonymous):

so i take the derivative of the given f(x) right ?

6 years ago
OpenStudy (anonymous):

yesss

6 years ago
OpenStudy (anonymous):

so i found the equation for the first given function f(x) = i/2x and got y= -1/2x-1/2. is that correct? can someone please confirm this

6 years ago
OpenStudy (anonymous):

sorry i meant the first function to be 1/2x not i/2x

6 years ago
OpenStudy (anonymous):

attach work

6 years ago
OpenStudy (anonymous):

okay so first: f(x)=1/2x the derivative of this is f'(x)= -1/2x^2 then i plugged in for x, F'(-1)= -1/2(-1)^2 and got -1/2 as my slope. then using the formula y-y1 = m(x-x1) I plugged in for m and finally got the equation y= -1/2x-1/2

6 years ago
OpenStudy (anonymous):

??

6 years ago
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