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?

Question

anonymous · Answer

just like the last one

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

yeah right

anonymous · Answer

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

anonymous · Answer

no don't think so

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

not sure where the 36 came from

anonymous · Answer

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

anonymous · Answer

ah right

anonymous · Answer

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