# `f(x) = kx^3, y = x + 1` Find k such that the line is tangent to the graph of the function.

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Expert Answers

kalau | Certified Educator

The tangent line must touch a point on the f(x) function.

Set both equations equal to each other.

`kx^3 = x+1`

Take the derivative of f(x) to find another relationship with k and x. The k is a constant, so do not eliminate it!

`f(x)= kx^3`

`f'(x)= 3kx^2`

Set this equal to the slope of the tangent line given, 1.

`1= 3kx^2`

Solve for k, since this is easier than solving for x.

`k = 1/(3x^2)`

Substitute the k back into the first equation.

`(1/(3x^2))x^3 = x+1`

`x/3 = x+1`

Multiply three on both sides.

`x= 3x+3`

`-2x =3`

`x= -3/2`

Re-substitute the x back to the first equation to solve for k.

`k(-3/2)^3 = -3/2+1`

`k(-27/8) = -1/2`

`k=1/2 * 8/27 = 4/27`