Question

for the given function, find the points on the graph at which the tangent line has slope 1.

\[y=\frac{ 1 }{ 3 }x^3-2x^2+4x+1\]

Answer 1

derive the function and solve for the derivative = 1

Answer 2

so would it just be y'=x^2

Answer 3

ye, so i guess its +1 and -1

Answer 4

but that is not one of my choices.
my choices are

A. (1,3) and (3,4)
B. (0,3) and (3,3)
\[C.(1,\frac{ 10 }{ 3 }) and (3,4)\]
\[D. (0,\frac{ 10 }{ 3 })and (3,4)\]

Answer 5

\[y=-\frac{1}{3}x^3-2x^2+4x+1\\
y'=-x^2-4x+4\]

Answer 6

ok so I understand where
\[y=\frac{ 1 }{ 3 }x^3-2x^2+4x+1\]
=
\[y'=x^2-4x+4\]
but how do I get the points from there?

Answer 7

got it thanks the answer is C

Answer 8

Yeah, you were supposed to let y' = 1, like @P0sitr0n said.