if anybody could help with this that would be much appreciated:
Find the point on the graph f(x)=4/3x^3 where the tangent line is perpendicular to the line y=-16x+3
Thanks in advance
-Ryan
Remember two lines are perpendicular to one another if and only if their slopes are negative reciprocals of each other.Originally Posted by compuryan
Thus, $\displaystyle y=-16x+3$ means the perpendicular line has slope $\displaystyle 1/16$.
The slope of any given point on the graph $\displaystyle f(x)=\frac{4}{3}x^3$ is its derivative which is,
$\displaystyle f'(x)=4x^2$.
We need that to be $\displaystyle 1/16$ as explained in the first paragraph thus, $\displaystyle 1/16=4x^2$. Solving we have that $\displaystyle x=\pm\frac{1}{8}$. These are the two x-values on the graph where the tangent is perpendicular to this line.