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

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

Answer 2

Find points where \[\frac{dy}{dx}=1\]

Answer 3

ok so would it be this?
\[1*\frac{ d }{ dx }=\frac{ 1 }{ 3 }x^3*\frac{ d }{ dx }-2x^2*\frac{ d }{ dx }+4x*\frac{ d }{ dx }+1\]

Answer 4

Why do you have d/dx multiplied into the equation like that?
I mean to find the derivative of y and set it equal to 0.

Answer 5

i dont know why i did that.