Question

Let f be the function given by f(x) = \[\sqrt{x^4-16x^2}\]
---- find the slope of the line normal to the graph of f at x=5

Answer 1

The first derivative is:\[\frac{2x(x^2-8)}{\sqrt{x^2(x^2-16)}}\]

Answer 2

That is the slope function. Replace x with 5 to find the slope of the tangent at x = 5

Answer 3

oh so i just plug it in and that gives me the slope?

Answer 4

\[\frac{2(5)(25-8)}{\sqrt{25(25-16)}}=\frac{170}{225}=\frac{34}{45}\]

Answer 5

That is the slope of the tangent. The slope of the perpendicular is:
\[\frac{-45}{34}\]

Answer 6

thank you!

Answer 7

yw