Question

can somebody help me with this question?
2: let f be the function given by :
f(x)=((x^4)-(16x^2))^(1/2)

Find the slope of the line normal to the graph of f at x = 5

Answer 1

anyone?

Answer 2

\[y=x^4-4x\]
\[y'=4x^3-4\]

Answer 3

The slope of the tangent at x = 5 is 4(5)^3-4 = 496

Answer 4

So the slope of the perpendicular is -1/496

Answer 5

where did u get 4x from?

Answer 6

\[\sqrt{16x ^{2}}=4x\]

Answer 7

Or is this the problem?
\[f(x)=\sqrt{x^4-16x^2}\]

Answer 8

that was the problem and it was 16x^2

Answer 9

So are you studying derivatives?

Answer 10

yea

Answer 11

Then take the derivative of the function.

Answer 12

i did then do wat next?

Answer 13

Replace x with 5 and evaluate.

Answer 14

That will be the slope of the tangent. The slope of the normal will be the negative reciprocal of the slope of the tangent.

Answer 15

i got 34/3 then the negative recip should be -(3/34) right?

Answer 16

yes

Answer 17

thxxx im finally done with my projext XD

Answer 18

I got 34/45 for the slope of the tangent.

Answer 19

let me redo it n c wat i did wrong

Answer 20

i typed everything into the caculator n i got 340/30---> 34/3

Answer 21

Your derivative must be wrong then. I put it into Wolf and got
\[\frac{2x(x^2-8)}{\sqrt{x^2(x^2-16)}}\]

Answer 22

it is i got (4x^3-32x)/ 2 sqaureroot x^4-16x^2

Answer 23

\[f(x)= \sqrt{x^{4}}-\sqrt{16x ^{2}}\] this was the original problem

Answer 24

So when I typed the problem and asked you if that was the problem and you said it was you really meant it wasn't?

Answer 25

nvm ur answer was correct mines wrong

Answer 26

wait so the slope is -(45/34) ?

Answer 27

yes

Answer 28

thx