Calculus - finding the minimal length between two functions

[Femme_physics]

My results are X1 = 4, X1 = -4
for its derivative = 0

So far so good?

 

[I like Serena]

My results are X1 = 4, X1 = -4
for its derivative = 0

So far so good?

Yes!

[edit]Actually your results are x₁ = 4, x₂ = -4 [/edit]

And now?

 

[Femme_physics]

Yes!

[edit]Actually your results are x₁ = 4, x₂ = -4 [/edit]

And now?

Now I plug it back into the formula for AB. I plug "4" because they're asking only for positive.

I get 1.5

Is that the answer, or do I have to plug that into the original function to find their y and then find the differences in the y?

In that case I get

\frac{1.5-2}{4} = -0.125

And
\frac{-4}{1.5} = -2.6666666

So the difference her is 2.79

 

[I like Serena]

Now I plug it back into the formula for AB. I plug "4" because they're asking only for positive.

I get 1.5

Is that the answer?

Yes!

or do I have to plug that into the original function to find their y and then find the differences in the y?

In that case I get

\frac{1.5-2}{4} = -0.125

And
\frac{-4}{1.5} = -2.6666666

So the difference her is 2.79

Nooooooo!
(You have calculated an y value. You're not supposed to then plug that in in f(x) as if it is an x value.)

 

[Mentallic]

You needed to find the difference in height between each function, and you found the x value where the height between them is minimal. We already have the function describing their difference (f(x)-g(x)) in height, so yes, just substituting x=4 to get y=3/2 is enough here.

 

[Femme_physics]

I'll have to do some backtracking to see how this all makes sense. I lost the string of the actual problem with all the math mishaps. Thank you all though! Much appreciate ILS for going for the kill! :)

 

[Mentallic]

I'll have to do some backtracking to see how this all makes sense. I lost the string of the actual problem with all the math mishaps. Thank you all though! Much appreciate ILS for going for the kill! :)

The basic idea of how to answer this question - without all the maths - is we first start with some function f(x) and another g(x), these functions don't intersect by looking at their graphs (we could also have proven this by setting f(x)=g(x) and shown there are no real solutions too, but it wasn't necessary here) and now to find the minimal distance in the y-direction between each function, we consider that the function f(x)-g(x) describes their differences in their height y at each point x, so for example if we took the value x=3, then f(3)=1/4 and g(3)=-4/3, which means their height difference is f(2)-g(2)=1/4-(-4/3)=19/12. So the function f(x)-g(x) is their height difference, and finding the minimal height means finding the lowest point on this graph (it will be above the y-axis, because we've already argued that the graphs don't intersect so f(x)-g(x)\neq0) and this is where calculus comes into it. Since when we find the derivative of a function, it tells us the gradient of the tangent at each x value, well we know the lowest point on the function has a tangent gradient of 0, so we find the derivative and set it to zero (this says "tangent gradient = 0") then solving for x, it will tell us the x-value where lowest point is. Now substituting x back into the derivative equation obviously yields 0, because we just solved the equality and found the value of x that makes the derivative 0. We need to substitute the x-value back into the function f(x)-g(x) which will give us the y-value, which is the height difference between the functions. This is it.