Answer
$x=1,- \frac{2}{3}$
Work Step by Step
$x+4y=10$
$y = -\frac{1}{4}x+\frac{10}{4}$
The slope of the line is $-\frac{1}{4}$. For the tangent line to be perpendicular to this line, the slope of the tangent line must therefore be $-1/\frac{-1}{4} = 4$
Taking the derivative of $f(x)$, we get $f'(x) = 6x^2-2x$
Equating $f'(x)$ to the desired slope, $4$, we get:
$6x^2-2x=4$
$6x^2-2x-4=0$
$x = 1,- \frac{2}{3}$
