## Calculus 10th Edition

$k = \frac{4}{27}$
To solve this problem, we need to identify point $(x,y)$ where both the function and the equation of the tangent line intersect. To do this, we can start by finding the first derivative of the function $f(x)$ using the Power Rule: $$f(x) = kx^{3}$$ $$f'(x) = 3kx^{3-1}$$ $$f'(x) = 3kx^{2}$$ Since the equation of the tangent line $y = x + 1$ gives us the slope $(m = 1)$, we can find the possible values of $x$: $$f'(x) = 1 = 3kx^{2}$$ $$\frac{1}{3k} = x^{2}$$ $$x = +/- \sqrt \frac{1}{3k}$$ To simplify, we can rewrite this in the following manner: $$x = +/- (\frac{1}{3k})^{\frac{1}{2}} = +/- (3k)^{- \frac{1}{2}}$$ Substituting this value in the original function to find $f(x)$, we get: $$f[(3k)^{- \frac{1}{2}}] = k *[ (3k)^{- \frac{1}{2}}]^{3}$$ which further simplifies to $$f[(3k)^{- \frac{1}{2}}] = k *[ (3k)^{- \frac{3}{2}}]$$ $$f[(3k)^{- \frac{1}{2}}] = k^{\frac{2}{2}} *[ (3)^{-\frac{3}{2}} * (k)^{- \frac{3}{2}}]$$ $$f[(3k)^{- \frac{1}{2}}] = k^{\frac{2}{2}} * (k)^{- \frac{3}{2}} * (3)^{-\frac{3}{2}}$$ $$f[(3k)^{- \frac{1}{2}}] = k^{-\frac{1}{2}} * 3^{-\frac{3}{2}}$$ $$f[(3k)^{- \frac{1}{2}}] = k^{-\frac{1}{2}} * 3^{-\frac{1}{2}} * 3^{-\frac{2}{2}}$$ $$f[(3k)^{- \frac{1}{2}}] = (3k)^{- \frac{1}{2}} * 3^{-1}$$ $$f[(3k)^{- \frac{1}{2}}] = \frac{1}{3}(3k)^{- \frac{1}{2}}$$ Notice that we've arrived at a solution that is written in terms of our previous solution to $x = (3k)^{- \frac{1}{2}}$. Therefore, we can state that the point of tangency in this entire exercise can be expressed as $(x, \frac{1}{3}x)$. As it stands, we can now input this information in the equation of the tangent line: $$y = x + 1$$ $$(\frac{1}{3}x) = x + 1$$ $$-\frac{2}{3}x = 1$$ $$x = -\frac{3}{2}$$ Finally, we can substitute $x$ for our previously calculated value: $$((3k)^{- \frac{1}{2}}) = -\frac{3}{2}$$ $$[(3k)^{- \frac{1}{2}}]^{-2} = [-\frac{3}{2}]^{-2}$$ $$3k = \frac{4}{9}$$ $$k = \frac{4}{27}$$