Determine a function and derivative from a given graph

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Homework Help Overview

The discussion revolves around determining the derivatives of functions based on provided graphs. The original poster is tasked with finding u'(1) and v'(5) using the Product Rule and Quotient Rule, but expresses confusion regarding the interpretation of the graphs and the functions they represent.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the identification of slopes from the graphs and the creation of linear functions for the piecewise segments of f(x). Questions arise about how to apply the product and quotient rules without explicit function definitions. Some participants suggest that values can be read directly from the graphs.

Discussion Status

There is ongoing exploration of the concepts of derivatives and tangents, with some participants providing guidance on how to approach the problem using the rules of calculus. The original poster has expressed some understanding but continues to seek clarity on the application of these concepts in the context of the graphs.

Contextual Notes

The original poster is working independently and has noted a lack of formal instruction, which may contribute to their confusion. There is also mention of the need to connect mathematical concepts to real-life situations.

ElectricRay
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Homework Statement


I have given two graphs which i try to show in the picture here. The question into find u'(1) and v'(5)

20170425_195420.jpg

Homework Equations


So the relevant equations here are the Product Rule and the Quotient Rule, which I know and is not the big problem here. I think (but imnot sure) the issue here is to find the primitive functions. But these graphs change ? For example the function f(x) has 3 different equations IMHO see x=-2 till 0 and x=0 till 2 and x=2 till 7 (at least that's what visible on the given graph)

The Attempt at a Solution


Well I could determine the slopes and create 3 linear functions for f and two linear functions for g.
the slope for f(x) in interval x = 0 till 2:

m = (4-0)/ (2-0) = 2
y(2) = 2*2 + b = 4
b = 0

So I can say that f = 2x+0, All great but i think I am completely lost. Main thing I don't understand the question?
The derivative is the slope (right?), but if I don't have the functions how can I determine the slope, well I just did that so what should I do now. Or is my answer m= 2 basically u'(1)? This seems like but it would also be for u'(1.5).

Can anybody tell me where I miss the point? And next when would this happen in a practically situation? I like to connect the things to real life situations.

Thanks in advance and apologizes if someone can't understand my confusion. I have calculus on school but they try to teach me everything with about 50 exercises so I bought A big calculus book and doing the exercises alone as I'm doing distance learning no teachers around.
 
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You should write down the product rule and put the x value of interest into it. You should be able to easily determine the value of everything needed by the product rule. The same is true for the quotient rule.

PS. The slope you calculated is correct. So is the equation you calculated. It was not really necessary to determine the equation -- you can read the function values straight from the graph.
 
Hi Ray:

What do you know about the slope and the angle of tangent to a function's curve with respect to an axis?
What do you know about the relationship of a tangent to distances along the x and y axis?

Hope that helps.

Regards,
Buzz
 
FactChecker said:
You should write down the product rule and put the x value of interest into it. You should be able to easily determine the value of everything needed by the product rule. The same is true for the quotient rule.

I think i can plug in values only if I have the functions

Buzz Bloom said:
Hi Ray:

What do you know about the slope and the angle of tangent to a function's curve with respect to an axis?
What do you know about the relationship of a tangent to distances along the x and y axis?
Buzz

What I know is as follows:
The slope is the point at which the tanget "touches" the curve a certain point. If I take (for example) the derivative of the function f(x) = x^2 I know that the slope will be at x=3. f'(3) = 2x so 2*3 = 6. This means that the slope will be 6 when x = 3. This part I understand...I think. Now I can say also that f(3) = 3^2 = 9 so i have the x,y coördinates and the slope which could help me create a linear function. But I have the feeling somthing in this whole derivate stuff is not completely clear to me ...yet :)
Nevertheless this is awesome stuff.
 
Hi Ray:

So for doing the u'(1) part of the problem, you should be able to write down f(1), f'(1), g(1), and g'(1). From that you can use the product formula to calculate u'(1).
Then for v'(5), you can write down the corresponding four values for x=5, and then use the quotient formula.
Are you OK with that?

Regards,
Buzz
 
ElectricRay said:
I think i can plug in values only if I have the functions
You are making it harder than you have to. You already read several function values straight from the graph to get the slope, m=2, of f(x). You can read the value of f(1) from the graph.

PS. You still haven't written down the product rule. Write it down here and fill in as much as you can.
 
Thread moved. @ElectricRay, please post questions involving derivatives in the Calculus & Beyond section, not the Precalculus section.
 
ElectricRay said:
think i can plug in values only if I have the functions
You have the graphs of the functions, so the formulas for the functions aren't necessary. If you had to, you could come up with piecewise definitions for f and g from your graph, but such isn't necessary in this problem.
 
Mark44 said:
Thread moved. @ElectricRay, please post questions involving derivatives in the Calculus & Beyond section, not the Precalculus section.
My apologizes I made a real mistake in selecting.

No regarding all the other help. I think I solved as you guys told me I could plug in the numbers.

The product rule part:
u(x) = f'(x) * g(x) + f(x)* g'(x)
u(1) = 2 * 1 + 2 * -1 = 0

The Quotien rule part:
For the slopes I have found
f'(x) = -1/3 and g'(x) = 2/3
Thus:
v(x) = [f'(x)*g(x) - f(x)*g(x)] / g(x)^2
v(5) = [-1/3 * 2 - 2/3 * 3] / 4
v(5) = -2/3

I that's all indeed i was making it myself to complicated.
Thanks for the advises
 
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