Use your knowledge of graphs of functions and their derivative to find the value of...

john458776

john458776

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I'm practicing for my upcoming test for advanced calculus. I have used my knowledge of derivative to find the value of the g(x)/k(x) but I'm not sure whether what I have applied to this problem is correct or accurate. Can you guide me. If I made mistake somewhere please tell me how can i approach this problem.

This is how I approached it using my knowledge of graphs.
20210515_191336.jpg
 
john458776 said:
View attachment 27221

I'm practicing for my upcoming test for advanced calculus. I have used my knowledge of derivative to find the value of the g(x)/k(x) but I'm not sure whether what I have applied to this problem is correct or accurate. Can you guide me. If I made mistake somewhere please tell me how can i approach this problem.

This is how I approached it using my knowledge of graphs.
View attachment 27222
Click to expand...
How did you conclude g'(0) = f(0) and k'(0) = h(0)?
 
Subhotosh Khan said:
How did you conclude g'(0) = f(0) and k'(0) = h(0)?
Click to expand...
The graph of g(x) is a cubic function so the graph of g'(x) would be a parabola graph since f(x) is a parabola graph I assume that it's the graph of g'(x).
I dont know if the knowledge i have is incorrect. does it make sense?
 
Look at the graph of g(x) and decide what the slope of the tangent line is at x=0. An advanced calculus student should know that will be g'(0).
The derivative of k(x) at k=0 is the slope of the tangent line of k(x) at k=0.

I do agree that the derivative of a cubic is a quadratic. The issue is will any quadratic work??

To me, it is extremely clear that g'(0) = k'(0). This should help you.
 
k(x) is not a parabola. It looks more like a sine graph.
g(x) looks more like a cubic.
1621154438129.png
In any case, it doesn't matter what the graphs are. The important thing is to compare the gradients of the two graphs at 0.
If you take a ruler and draw in the tangent to each graph at 0, you'll see that [MATH]g'(0) > k'(0)[/MATH] (and both are non-zero). Therefore we can conclude that [MATH]\hspace2ex \lim \limits_{x \to 0} \frac{g(x)}{k(x)} >1[/MATH]
 
lex said:
k(x) is not a parabola. It looks more like a sine graph.
g(x) looks more like a cubic.
View attachment 27232
In any case, it doesn't matter what the graphs are. The important thing is to compare the gradients of the two graphs at 0.
If you take a ruler and draw in the tangent to each graph at 0, you'll see that [MATH]g'(0) > k'(0)[/MATH] (and both are non-zero). Therefore we can conclude that [MATH]\hspace2ex \lim \limits_{x \to 0} \frac{g(x)}{k(x)} >1[/MATH]
Click to expand...

lex, his approach is not for the same problem, but for a similar problem
 
lex said:
k(x) is not a parabola. It looks more like a sine graph.
g(x) looks more like a cubic.
View attachment 27232
In any case, it doesn't matter what the graphs are. The important thing is to compare the gradients of the two graphs at 0.
If you take a ruler and draw in the tangent to each graph at 0, you'll see that [MATH]g'(0) > k'(0)[/MATH] (and both are non-zero). Therefore we can conclude that [MATH]\hspace2ex \lim \limits_{x \to 0} \frac{g(x)}{k(x)} >1[/MATH]
Click to expand...
And lex, your conclusion here is correct, the limit should be greater than 1

and from the graphs in the main question, it makes sense the answer is [MATH]\frac{3}{2}[/MATH]
 
nasi112 said:
His main question had a sine function and a cubic function

but his approach was done on a parabola function and an exponential function
Click to expand...
I'm not sure that they haven't just changed their opinion about the original question! We must wait and see.
 
lex said:
k(x) is not a parabola. It looks more like a sine graph.
g(x) looks more like a cubic.
View attachment 27232
In any case, it doesn't matter what the graphs are. The important thing is to compare the gradients of the two graphs at 0.
If you take a ruler and draw in the tangent to each graph at 0, you'll see that [MATH]g'(0) > k'(0)[/MATH] (and both are non-zero). Therefore we can conclude that [MATH]\hspace2ex \lim \limits_{x \to 0} \frac{g(x)}{k(x)} >1[/MATH]
Click to expand...
So you think [MATH]3/2[/MATH] is correct since it is greater than 1.

I tried to find the function of g(x) and k(x) but I dont think they are correct. I think [MATH] g(x)=x^3[/MATH] and [MATH]k(x)=sin(x)[/MATH] But this gives me zero.

I tried again [MATH] g(x)=tanx[/MATH] and [MATH]k(x)=sin(x)[/MATH] But this gives me [MATH]5/2[/MATH].

I'm really confused how I will approach this question.
 
nasi112 said:
And lex, your conclusion here is correct, the limit should be greater than 1

and from the graphs in the main question, it makes sense the answer is [MATH]\frac{3}{2}[/MATH]
Click to expand...
You think [MATH]\frac{3}{2}[/MATH] is correct according to Lex conclusion.
 
Jomo said:
Look at the graph of g(x) and decide what the slope of the tangent line is at x=0. An advanced calculus student should know that will be g'(0).
The derivative of k(x) at k=0 is the slope of the tangent line of k(x) at k=0.

I do agree that the derivative of a cubic is a quadratic. The issue is will any quadratic work??

To me, it is extremely clear that g'(0) = k'(0). This should help you.
Click to expand...
g(x) & k(x) intersect each other at x = 0. Those aren't tangential to each other. It is clear though that g'(x) > k'(x) > 0.
 
nasi112 said:
Try this

[MATH]g(x) = \frac{5x^3}{3} + 3x[/MATH]
[MATH]k(x) = sin(2x)[/MATH]
[MATH]\lim_{x\to0} \frac{g(x)}{k(x)} = \ ?[/MATH]
Click to expand...
[MATH]\lim_{x\to0} \frac{g(x)}{k(x)} = 3/2[/MATH]
How did you come up with the functions?
 
john458776 said:
I'm really confused how I will approach this question.
Click to expand...
There are 2 ways to go, rather than guessing what the original functions were.
1. Observe that [MATH]g'(0) > k'(0)[/MATH] (and both are non-zero). Therefore, conclude that [MATH]\hspace2ex \lim \limits_{x \to 0} \frac{g(x)}{k(x)} >1[/MATH]or
2. Assume that the little dots on the grid represent the same units on the horizontal and vertical directions.
Take a ruler and draw in the tangent to each graph at 0. Work out the gradients of these 2 lines. They will represent g'(0) and k'(0). Then write down
[MATH]\frac{g'(0)}{k'(0)}[/MATH].
That's all the question is looking for; not what the original functions were.
 
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