Question

last one @jim_thompson5910

Answer 1

I have no idea how to do this

Answer 2

\[\Large h(x) = \frac{f(x)}{x}\]

are you able to compute what `h ' (x) ` would be? (in terms of x and f(x))

Answer 3

hint: quotient rule

Answer 4

\[h'=\frac{ f'x-f }{ x }\]

Answer 5

how can I solve that if I don't know the equation of f(x)

Answer 6

sorry, bottom is x^2

Answer 7

we know f=3 and f(3)=10

Answer 8

so, h'=(3f'-10)/9

Answer 9

now, f'=-1 from graph, so h'=-13/9

Answer 10

you don't need to know f(x) itself

you just need to use the quotient rule to get

\[\Large h(x) = \frac{f(x)}{x}\]

\[\Large h \ '(x) = \frac{f \ '(x)*x-f(x)*1}{x^2}\]

\[\Large h \ '(x) = \frac{f \ '(x)*x-f(x)}{x^2}\]

does that make sense?

Answer 11

ah, I see.

Answer 12

I forgot they gave you the values for f(3)

Answer 13

so y=-13/9x+b

Answer 14

x=3, y=h=10/3

Answer 15

now solve for b and you are done

Answer 16

thank you so much! You're amazing

b = 23/3

Answer 17

Thank you both for your help. Thanks for sticking with me this far @jim_thompson5910

Answer 18

yes you'll then replace x with 3
f(3) = 10
so

\[\Large h \ '(x) = \frac{f '(x)*x-f(x)}{x^2}\]

\[\Large h \ '(3) = \frac{f '(3)*3-f(3)}{3^2}\]

\[\Large h \ '(3) = \frac{3*f \ '(3)-10}{9}\]

the question is: how to find f ' (3)? This value is not given to us. But we can approximate it by following these steps


step 1) mark the points A,B,C (see attached)

step 2) find the slope of AB

step 3) find the slope of BC

step 4) average the two slopes (AB and BC)

this will give you an approximate value of f ' (3)

Answer 19

ideally we get as close to x = 3 as possible, but this is as close as we can get

Answer 20

Awesome. Again, thank you so much. :)

Answer 21

no problem

Answer 22

Actually, isn't the value of f'(x) given at the point (3,-1) ?

Answer 23

oh wow, now it's my turn to mix up f and f '

my bad lol

Answer 24

yeah they provided (3,-1) on f '

so f ' (3) = -1

Answer 25

so, would h ' = -13/9

Answer 26

h ' (3) = -13/9, yes

Answer 27

and that would be the equation for the line tangent to h(x) at x = 3?

Answer 28

that's the slope of the tangent line on h(x) at x = 3

Answer 29

ah, so plug into equation for a line. got it

Answer 30

thx again :)

Answer 31

oh, and would you plug in 10 for y?

Answer 32

h(x) = f(x)/x

h(3) = f(3)/3 ... replace every x with 3

h(3) = 10/3 ... replace f(3) with 10

the y coordinate is actually 10/3

Answer 33

ok, thanks. Again, lol. I promise you can leave now.

Answer 34

you already found it above when you solved

\[\Large y = mx+b\]

\[\Large \frac{10}{3} = -\frac{13}{9}*3 + b\]

for b to get b = 23/3

Answer 35

so your tangent line equation is
\[\Large y = -\frac{13}{9}x + \frac{23}{3}\]

Answer 36

oh, yeah...duh. How did I miss that?