Comparing Derivatives: Solving for g'(x) and f'(x) with Two Given Points

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In summary, the conversation discusses the comparison of derivatives of two differentiable functions, f(x) and g(x), in the interval (2,4). Through the use of the mean value theorem and considering the functions h(x)=g(x)-4f(x) and v(x)=3g(x)-4f(x), it is determined that the statement D, g'(x)=4f'(x) for at least one x in (2,4), is true. Option B is a red herring and a counter-example can be used to show that v'(x) is never zero in (2,4).
  • #1
Saitama
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Homework Statement


Let ##f(x)## and ##g(x)## be two differentiable function in R and f(2)=8, g(2)=0, f(4)=10 and g(4)=8 then

A)##g'(x)>4f'(x) \forall \, x \, \in (2,4)##

B)##3g'(x)=4f'(x) \, \text{for at least one} \, x \, \in (2,4)##

C)##g(x)>f(x) \forall \, x \, \in (2,4)##

D)##g'(x)=4f'(x) \, \text{for at least one} \, x \, \in (2,4)##


Homework Equations





The Attempt at a Solution


How am I to compare the derivatives with only two points? I really don't know where to start with this. Just a wild guess, do I need to apply the mean value theorem?
 
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  • #2
Hi Pranav-Arora! :smile:

Hint: consider g - 4f and 3g - 4f :wink:
 
  • #3
tiny-tim said:
Hi Pranav-Arora! :smile:

Hint: consider g - 4f and 3g - 4f :wink:

I don't see how does it help. :confused:

Let ##h(x)=g(x)-4f(x)##. Then ##h(4)=h(2)=-32##. This suggests that h' is zero somewhere in (2,4). What should I do now?
 
  • #4
Pranav-Arora said:
This suggests that h' is zero somewhere in (2,4).

yup! :smile:

and h' = g' - 4f' :wink:
 
  • #5
tiny-tim said:
yup! :smile:

and h' = g' - 4f' :wink:

Ah yes, then its D. Thank you tiny-tim! :smile:

Let v(x)=3g(x)-4f(x). How do I prove that v'(x) is never zero in (2,4)? :confused:
 
  • #6
Pranav-Arora said:
ALet v(x)=3g(x)-4f(x). How do I prove that v'(x) is never zero in (2,4)? :confused:

B is just there to confuse you! :wink:

Anyway, you'd only need to show that it can be never-zero …

you should be able to sketch a counter-example. :smile:
 
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  • #7
tiny-tim said:
B is just there to confuse you! :wink:

Anyway, you'd only need to show that it can be never-zero …

you should be able to sketch a counter-example. :smile:

Okay, I understand, thank you once again! :)
 

1. What is the purpose of comparing derivatives?

Comparing derivatives allows scientists to analyze the rate of change of a function or system. This can provide insights into how the system is evolving or how different variables affect the function.

2. How do you compare derivatives?

To compare derivatives, you can graph the functions and observe their slopes at different points, or you can calculate the derivatives using the rules of differentiation.

3. What are some real-world applications of comparing derivatives?

Comparing derivatives is used in various fields such as physics, economics, and engineering. For example, it can be used to analyze the acceleration of an object, the rate of change of stock prices, or the growth rate of a population.

4. What are the key differences between derivatives and anti-derivatives?

Derivatives represent the instantaneous rate of change of a function, while anti-derivatives represent the original function before differentiation. Additionally, derivatives result in a single value, while anti-derivatives have a constant of integration.

5. What are the common mistakes to avoid when comparing derivatives?

Some common mistakes to avoid when comparing derivatives include not understanding the chain rule, forgetting to use the product or quotient rule, and not simplifying the derivative before comparing. It is also important to check for points of discontinuity or undefined values when comparing derivatives.

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