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

lex said:
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.
Click to expand...
I used method 2 to approach it, that's how I found out its 3/2 but I was not sure about it, it's my first time learning advanced calculus.

Thanks.
 
john458776 said:
[MATH]\lim_{x\to0} \frac{g(x)}{k(x)} = 3/2[/MATH]
How did you come up with the functions?
Click to expand...
you need the derivative to be a parabola that has a minimum at [MATH]g'(0) = 3[/MATH], then

[MATH]g'(x) = x^2 + 3[/MATH]
I used [MATH]5x^2 + 3[/MATH], just to make the graph look more similar to your graph. Both of them is valid.

integrate anyone of them to get [MATH]g(x)[/MATH]
Same thing for

[MATH]h'(x) = 2\cos(x)[/MATH], you need a cosine function that has maximum at [MATH]h'(0) = 2[/MATH]
when you integrate this function you will get [MATH]h(x) = 2\sin(x)[/MATH], but we need the height of the function to be [MATH]1[/MATH], instead of 2

then

Use [MATH]h'(x) = 2\cos(2x)[/MATH], instead of [MATH]h'(x) = 2\cos(x)[/MATH]
 
My eye sight must be worse than I thought as the tangent line of g(x) and k(x) at x=0 surely look the same to me.
 
@Jomo
Yes, it’s hard to be sure just by eye!
To convince yourself, use a ruler and you’ll see the difference.
(Also plot a graph where the two gradients are 1 and you’ll see the difference).
 
You must log in or register to reply here.
Top