Question

Let f(x) =sqrt[3] x. The equation of the tangent line to f(x) at x = 8 is y =? Using this, we find our approximation for sqrt[3] {8.2} is?

Answer 1

just like the last one

Answer 2

i was following the same format but i keep going wrong somewhere

Answer 3

\[f(x)=\sqrt[3]{x}\] right ?

Answer 4

i know you take the derivative of ∛(x) which is 1/ 3∛(x)^2 then plug 8 into x to find the slope

Answer 5

yeah right

Answer 6

and i calculated the slope would be 1/36 so the equation i got was y-0=(1/36)(x-8)

Answer 7

no don't think so

Answer 8

\[f'(x)=\frac{1}{3\sqrt[3]{x^2}}\]

Answer 9

\[f'(8)=\frac{1}{3\sqrt[3]{8^2}}=\frac{1}{3\times 2^2}=\frac{1}{12}\]

Answer 10

not sure where the 36 came from

Answer 11

because i thought the whole bottom was squared but it was actually only the x

Answer 12

ah right

Answer 13

the equation should be y-0=1/12(x-8), correct?