Question

f(x) = square root ( 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

f(x) = sqrt ( x^2 (x+4) (x-4) )
domain is all real numbers >= 4

Answer 2

(at least, that's the domain if you are not yet doing complex numbers -- are you doing complex numbers yet?)

Answer 3

i dont need to find complex numbers for this

Answer 4

i got -4 to be a domain too

Answer 5

No, what I mean is
- if you are considering only real-valued functions, then the domain is all real numbers >= 4 (I guess this is the case)
- you will learn later that for complex-valued functions, the domain becomes all real numbers (you're probably not there yet).

Answer 6

So I think the answer you want is the one I gave at the top.

Answer 7

how do you do part c ?

Answer 8

If you are using the set of real numbers then the domain is

\[x \le -4, x \ge 4\]


for the derivative

\[f'(x) = 1/2(x^4 - 16x^2)^{-1/2} \times (4x^3 - 32x)\]

find the gradient of the tangent (m) at x =5 by substituting x = 5. Then the the gradient of the normal is -1/m

Answer 9

Once you get (b), which is a small mess but doable, just evaluate f'(5) and take the negative reciprocal.

Answer 10

with the domain.. be careful and (-4)^ - 16(-4)^2 gives the same answer as (4)^4 - 16(4)^2

Answer 11

so the domain is \[x \le -4, x \ge 4\]

Answer 12

f(x) = square root ( x^4 -16x^2)

Domain: x^4 -16x^2 ≥ 0
-> x^2 ( x^2 - 16) ≥ 0 -> x^2 ≥ 16
=> x <= -4, or x ≥ 4

f'(x) = (2x^3 - 16x) / square root ( x^4 -16x^2)

c) Find the slope of the line normal to the graph of f at x = 5
Tangent line slope:
f'(5) = (2 * 5^3 - 16 * 5)/ sqrt ( 5^4 - 16* 5^2) = 170/15 = 34/3

normal line slope m = -3/34