Question

calculus question....
let f(x) = (x^2+4)/x
write an equation of the normal line to the graph of f at the point where x=2

Answer 1

\(f(2)=4\) so the point is \((2,4)\) now you need the slope
did you find the derivative?

Answer 2

i suppose you could use the quotient rule, but it is easier to divide and get
\[f(x)=x+\frac{4}{x}\] so \(f'(x)=1-\frac{4}{x^2}\) and
\[f'(2)=1-1=0\] hmm is this a trick question?

Answer 3

no matter, tangent line is horizontal, normal line is vertical, and vertical line through \((2,4)\) is \(x=2\)

Answer 4

i got as the derivative of the f(x) x-4

Answer 5

it is
\[f(x)=\frac{x^2+4}{x}\]?

Answer 6

yes so the derivative is x-4
and i got the equation for the tangent line at x=1 but unsure how to find the normal line at x=2

Answer 7

hold on
the derivative is not \(x-4\)

Answer 8

\(x-4\) is the derivative of \(\frac{1}{2}x^2-4x\) not \(\frac{x^2+4}{x}\)

Answer 9

i am not sure what rule you used, but you can either use the quotient rule, or more easily divide and take the derivative piece by piece

Answer 10

i used quotient rule.... \[gf'-fg'/g^2\]

Answer 11

i got y-4=-1/3(x-2)

Answer 12

ok lets get the derivative correct first, then we can find the slope

Answer 13

\[\frac{gf'-fg'}{g^2}\] with
\[f(x)=x^2+4,f'(x)=2x,g(x)=x,g'(x)=1\] you get
\[\frac{x\times 2x-(x^2+4)}{x^2}=\frac{x^2+4}{x^2}\]

Answer 14

damn typo it is
\[\frac{x^2-4}{x^2}\] sorry

Answer 15

its a test review.... :( im gunna fail lol

Answer 16

although it was easier to compute the derivative of
\[f(x)=x+\frac{4}{x}\] then \[f'(x)=1-\frac{4}{x^2}\]

Answer 17

nah, just some practice, but look the only way you are going to get a derivatie that is a nice line line \(x-4\) is if you start with a parabola like \(\frac{1}{2}x^2-4x\)

Answer 18

if you have a quotient expect to get a numerator and a denominator for the derivative
the derivative of a rational function will be another rational function, not a line

Answer 19

mm i got it its 2x-4 cause the top x2 and bottom x2 cancel out.

Answer 20

whoa nelly stop cancelling!

Answer 21

don't expect to cancel ever. there is rarely cancelling in the quotient rule, and in any case \(x^2\) is not a factor of the numerator,
careful with the algebra, that will mess you up for sure

Answer 22

but don't forget, your job is to evaluate the derivative, so no need to "simplify" and make a mistake
just plug in 2 and see what you get

Answer 23

its the only way i see it working lol the tops neg and bottom is positive... because f' is 2 and g' is 1.

Answer 24

\(f'(x)=2x\) not \(2\)

Answer 25

ok hold on. i think i got. thanks :)

Answer 26

if you are taking a test, remember that the derivative of a rational function is another rational function, not a line or any other polynomial
only "combine like terms" in the numerator, do not cancel anything

Answer 27

yw

Answer 28

ok so wait i got the derivative. so back to original question how do you solve for the slope. and new question how do you get the points where f has a horizontal tangent?