Question

f(x)= (square root(x^4-16x^2)
a. Find the domain of f.
b.Find F'(x)
c.find te slope of the line normal to the graph of f at x=6

Answer 1

domain is when that inside of that radical is zero or positive
so you need to solve
\[x^4-16x^2 \ge 0\]

Answer 2

how do i do that?

Answer 3

\[x^2(x^2-16) \ge 0\]
let \[g(x)=x^2(x^2-16)\]
g is zero when x=-4,0,4

So I would draw a number line and plot these numbers and test before and after to see if g>0


-----|-----|------|-----
-4 0 4
g(-5) | g(-1)| g(1) | g(5)
=+(+)|=+(-)| =+(-)|=+(+)
=+ =- =- =+

So the domain is when g=0 or when g>0

Answer 4

so youwol gtx= 0, 4 ,-4 so yor dmai wod e all numbers such ta x cnnot
equal t 0,4,-4?

Answer 5

that happens on the intervals and at the numbers:
\[(-\infty,-4] U \{0\} U [4,\infty)\]

Answer 6

now to find f' you need to use chain rule!

Answer 7

\[f'=\frac{1}{2}(x^4-16x^2)^\frac{-1}{2}(x^4-16x^2)'\]

Answer 8

nd hn he 1/2 is multplid b the (x^4-16x^2)?

Answer 9

i don't understand the question

Answer 10

ok uno how is

Answer 11

ok you want to multiply
the order is commutative so you can do that

Answer 12

\[f'=\frac{2x^3-16x}{\sqrt{x^4-16x^2}} \]