Answer
$k = 27$
Work Step by Step
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: $$f(x) = kx^{4}$$ $$f'(x) = 4kx^{4-1}$$ $$f'(x) = 4kx^{3}$$ Since the equation of the tangent line $y = 4x - 1$ gives us the slope $(m = 4)$, we can proceed to find the possible values of $x$: $$f'(x) = 4 = 4kx^{3}$$ $$1 = kx^{3}$$ $$[\frac{1}{k}]^{\frac{1}{3}} = [x^{3}]^{\frac{1}{3}}$$ $$k^{-\frac{1}{3}} = x$$ Substituting this value into the original function to find $f(x)$, we get: $$f(k^{-\frac{1}{3}}) = k[k^{-\frac{1}{3}}]^{4}$$ $$f(k^{-\frac{1}{3}}) = k^{\frac{3}{3}} * [k^{-\frac{4}{3}}]$$ $$f(k^{-\frac{1}{3}}) = k^{-\frac{1}{3}}$$ Notice that the point of tangency $(x, f(x))$ in this case is $(k^{-\frac{1}{3}}, k^{-\frac{1}{3}})$ and can be written as $(x, x)$ for simplicity. Therefore, when plugging these values into the equation for the tangent line provided: $$y = 4x - 1$$ $$(x) = 4(x) - 1$$ $$1 = 3x$$ $$\frac{1}{3} = x$$ Substituting the value of $x$, we finally get: $$\frac{1}{3} = (k^{-\frac{1}{3}})$$ $$[3^{-1}]^{-3} = (k^{-\frac{1}{3}})^{-3}$$ $$3^{3} = k$$ $$27 = k$$
