Question

find the equation of the tangent line at the point indicated...

Answer 1

f(x)=\[\log_{2}(7x+x ^{-7}) \] at x=1

Answer 2

\[f'(x)=\frac{7-\frac{7}{x^8}}{(7x+\frac{1}{x^7})\ln(2)}\]

Answer 3

please show all steps so i understand :)

Answer 4

\[f(x)=\frac{\ln(7x+x^{-7})}{\ln(2)} => f'(x)=\frac{1}{\ln(2)} \cdot \frac{7-7x^{-8}}{7x+x^{-7}}\]
\[f'(1)=\frac{1}{\ln(2)} \cdot \frac{7-7(1)^{-8}}{7(1)+(1)^{-7}}=\frac{1}{\ln(2)} \cdot \frac{7-7}{7+1}=0\]

Answer 5

looks like
\[f'(1)=0\] unless i made a mistake

Answer 6

so the equation of the tangent line would be y=0?

Answer 7

nope, i guess i was right according to myininaya. so your point is
\[(1,3)\] slope is 0, equation is
\[y=3\]

Answer 8

no that is the slope

Answer 9

where did the 3 come from?

Answer 10

sorry i filled in x variable in wrong equation.

Answer 11

what?

Answer 12

where did the 3 come from?

Answer 13

\[f(1)=?\]

Answer 14

that is what i said.

Answer 15

\[2^{blank}=8\]

Answer 16

you need a point and a slope. slope is 0, point is
\[(1,f(1))\]

Answer 17

what is blank = to?

Answer 18

in this case
\[f(1)=\log_2(7\times 1+\frac{1}{1^7})=\log_2(8)\]

Answer 19

ok

Answer 20

another random question... is the derivative of f(x) =e^12 12 or 0? i get confused sometimes

Answer 21

0

Answer 22

thanks guys!

Answer 23

because
\[e^{12}\] is a number

Answer 24

thats what i thought!