Question

need help on the attachment

Answer 1

they want you to find the equation of the tangent line

Answer 2

wants the equation for the line right?

Answer 3

yes

Answer 4

take the derivative, and then find
\[f'(1)\] as your slope. use point-slope formula

Answer 5

I got 3/2(2)^(1/2) for the slope

Answer 6

\[f'(x)=\frac{2}{\sqrt[3]{(6x+2)^2}}\]

Answer 7

oops I actually put square root of 1/2 instead 1/3

Answer 8

\[f(x)=(6x+2)^{\frac{1}{3}}\]
\[f'(x)=\frac{1}{3}(6x+2)^{-\frac{2}{3}}\times 6\]

Answer 9

now it should be easy because
\[f'(1)=\frac{2}{\sqrt[3]{8^2}}\]

Answer 10

aka
\[\frac{1}{2}\]

Answer 11

slope is 3/4

Answer 12

really? let me read it again

Answer 13

oops I meant 1/2

Answer 14

whew

Answer 15

and point is
\[(1,2)\] and point slope formula will give it to you

Answer 16

where did you get 2 from

Answer 17

if
\[x=1\] then
\[y=f(1)=(6\times 1+2)^{\frac{1}{3}}=8^{\frac{1}{3}}=\sqrt[3]{8}=2\]

Answer 18

oh, ok

Answer 19

answer y=(1/2x)+(3/2)

Answer 20

i believe you

Answer 21

Thanks so much!

Answer 22

yw