Question

If g(x) = 8 − x3, find g'(0) and use it to find an equation of the tangent line to the curve y = 8 − x3 at the point (0, 8)

Answer 1

is it
\[g(x)=8-x^3\]?

Answer 2

Yes

Answer 3

you got the derivative?

Answer 4

I can only come up with 0.

Answer 5

yeah it is \(g'(x)=-3x^2\) making \(g'(0)=0\)

Answer 6

slope of the tangent lines is zero, line is horizontal

Answer 7

So the equation is straight line ?

Answer 8

all lines are straight (unlike my friends)

Answer 9

it is horizontal i.e. \(y=8\)

Answer 10

Thank you,