Question

Please help

\[f(x)=\sqrt{x ^4-16x ^{2}}\]
Let f be the function given by a) Find the domain of f. b) Find f '(x) c) Find the slope of the line normal to the graph of f at x = 5

Answer 1

(x^4-16x^2)^(1/2)

x^4 - 16x^2 >= 0
x^4 >= 16x^2
x^2 >= 16
x >= 4
so the domain is [4, infinity)

Answer 2

derivative of (x^4-16x^2)^(1/2)

(1/2)(x^4-16x^2)^(-1/2)(4x^3-32x) by the chain rule

(4x^3-32x)/(2(sqrt(x^4-16x^2)))

Answer 3

to find the slope at x=5, plug 5 into the equation above

Answer 4

the domain is \[(-\infty,-4]\cup[0]\cup[4,\infty)\]

Answer 5

Thanks and do you plug 5 into the original equation or f prime?

Answer 6

plug it in to f prime to get the slope

Answer 7

and then what? Sorry

Answer 8

i got 34/3 for the slope

Answer 9

that's it. the slope at x=5 is 34/3

Answer 10

Oh alright thanks

Answer 11

note that is just the slope of the tangent line...not the normal line

Answer 12

How do you find the normal line then ?

Answer 13

the reciprocal?

Answer 14

if \(m\) is the slope of the tangent line then \(-\frac{1}{m}\) is the slope of the normal line.

Answer 15

negative reciprocal

Answer 16

so -3/34?

Answer 17

yes

Answer 18

thanks