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

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

Question

for the given function, find the points on the graph at which the tangent line has slope 1.

$$y=\frac{ 1 }{ 3 }x^3-2x^2+4x+1$$

p0sitr0n · Answer

derive the function and solve for the derivative = 1

anonymous · Answer

so would it just be y'=x^2

p0sitr0n · Answer

ye, so i guess its +1 and -1

anonymous · Answer

but that is not one of my choices.
my choices are

A. (1,3) and (3,4)
B. (0,3) and (3,3)
$$C.(1,\frac{ 10 }{ 3 }) and (3,4)$$
$$D. (0,\frac{ 10 }{ 3 })and (3,4)$$

anonymous · Answer

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

anonymous · Answer

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?

anonymous · Answer

got it thanks the answer is C

anonymous · Answer

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