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Use the graph below to complete the following where h(x) = f(x)g(x) and m(x) = f(x)/g(x) graph

  1. h'(x)
  2. m'(x)
  3. m'(2)
  4. h'(−7)
  5. h'(-5)

The first two problems are quite straightforward since we know that h(x) = f(x)g(x) and m(x) = f(x)/g(x).

1. h'(x) = f'(x)g(x) + f(x)g'(x)

2. m'(x) = (f'(x)g(x) - f(x)g'(x)) / (g(x))^2

Starting from number 3, I am not sure how I am supposed to do it. I know that by plugging-in the value, we get:

3. m'(2) = (f'(2)g(2) - f(2)g'(2)) / (g(2))^2

where g(2) = -1 and f(2) = 1.3 (not sure about f(2))

In this problem, how am I supposed to find f'(2) and g'(2)? I can't quite figure it out. Any help would be appreciated. Thanks!

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2  
Notice that $ \ f(x) \ $ is a straight line (thus, constant slope) on the interval $ \ 0 \ \le \ x \ \le \ 6 \ $ and $ \ g(x) \ $ has a horizontal tangent at $ \ x \ = \ 2 \ . $ Find the slope of the line for $ \ f(x) \ $ and you'll have all the information you need. Something similar will work for #4 ; beware on #5 : what can be said about $ \ f'(-5) \ $ ? –  RecklessReckoner Jan 12 at 22:10
    
@RecklessReckoner In this case, f'(2) = -4/3 ((y2- y1) / (x2 - x1)) and g'(2) = 0. Then the answer for #3 will be m'(2) = DNE because we get the expression divided by 0. Is this correct? For #5, h'(-5) = DNE because f'(-5) = DNE (sharp corner). –  KurodaTsubasa Jan 12 at 22:38
1  
Yes on #5: $ \ f(x) \ $ has no defined derivative at $ \ x \ = \ -5 \ , $ so neither will $ \ f(x) \ $ . On #3 , it's $ \ g'(x) \ $ that is zero at $ \ x = 2 \ $ , but $ \ g(2) \ $ is not zero (as you already noted)... –  RecklessReckoner Jan 12 at 22:44
    
@RecklessReckoner Oh, yes, my mistake. I mean the answer would be m'(2) = 4/3. Now I understand how I should approach these questions. Thank you very much! –  KurodaTsubasa Jan 12 at 23:05
    
Glad to hear that: this is a sort of question that sometimes appears on exams, using a graph, or tabular data, or specified functions. –  RecklessReckoner Jan 12 at 23:08

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