StudyGurl14
  • StudyGurl14
last one @jim_thompson5910
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
StudyGurl14
  • StudyGurl14
1 Attachment
StudyGurl14
  • StudyGurl14
I have no idea how to do this
jim_thompson5910
  • jim_thompson5910
\[\Large h(x) = \frac{f(x)}{x}\] are you able to compute what `h ' (x) ` would be? (in terms of x and f(x))

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

jim_thompson5910
  • jim_thompson5910
hint: quotient rule
caozeyuan
  • caozeyuan
\[h'=\frac{ f'x-f }{ x }\]
StudyGurl14
  • StudyGurl14
how can I solve that if I don't know the equation of f(x)
caozeyuan
  • caozeyuan
sorry, bottom is x^2
caozeyuan
  • caozeyuan
we know f=3 and f(3)=10
caozeyuan
  • caozeyuan
so, h'=(3f'-10)/9
caozeyuan
  • caozeyuan
now, f'=-1 from graph, so h'=-13/9
jim_thompson5910
  • jim_thompson5910
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?
StudyGurl14
  • StudyGurl14
ah, I see.
StudyGurl14
  • StudyGurl14
I forgot they gave you the values for f(3)
caozeyuan
  • caozeyuan
so y=-13/9x+b
caozeyuan
  • caozeyuan
x=3, y=h=10/3
caozeyuan
  • caozeyuan
now solve for b and you are done
StudyGurl14
  • StudyGurl14
thank you so much! You're amazing b = 23/3
StudyGurl14
  • StudyGurl14
Thank you both for your help. Thanks for sticking with me this far @jim_thompson5910
jim_thompson5910
  • jim_thompson5910
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)
1 Attachment
jim_thompson5910
  • jim_thompson5910
ideally we get as close to x = 3 as possible, but this is as close as we can get
StudyGurl14
  • StudyGurl14
Awesome. Again, thank you so much. :)
jim_thompson5910
  • jim_thompson5910
no problem
StudyGurl14
  • StudyGurl14
Actually, isn't the value of f'(x) given at the point (3,-1) ?
jim_thompson5910
  • jim_thompson5910
oh wow, now it's my turn to mix up f and f ' my bad lol
jim_thompson5910
  • jim_thompson5910
yeah they provided (3,-1) on f ' so f ' (3) = -1
StudyGurl14
  • StudyGurl14
so, would h ' = -13/9
jim_thompson5910
  • jim_thompson5910
h ' (3) = -13/9, yes
StudyGurl14
  • StudyGurl14
and that would be the equation for the line tangent to h(x) at x = 3?
jim_thompson5910
  • jim_thompson5910
that's the slope of the tangent line on h(x) at x = 3
StudyGurl14
  • StudyGurl14
ah, so plug into equation for a line. got it
StudyGurl14
  • StudyGurl14
thx again :)
StudyGurl14
  • StudyGurl14
oh, and would you plug in 10 for y?
jim_thompson5910
  • jim_thompson5910
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
StudyGurl14
  • StudyGurl14
ok, thanks. Again, lol. I promise you can leave now.
jim_thompson5910
  • jim_thompson5910
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
jim_thompson5910
  • jim_thompson5910
so your tangent line equation is \[\Large y = -\frac{13}{9}x + \frac{23}{3}\]
StudyGurl14
  • StudyGurl14
oh, yeah...duh. How did I miss that?

Looking for something else?

Not the answer you are looking for? Search for more explanations.