Minimum veritcal distance between two graphs.

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Homework Statement


Calculate the minimum vertical distance between the cubic y=x^3-x^2-4x+4 and the parabola y=-2x^2+16x-30, when x is positive.

Homework Equations


y=x^3-x^2-4x+4
y=-2x^2+16x-30

The Attempt at a Solution


I have no idea at all how to work this out, so perhaps someone could walk me through it?
 
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Made an attempt:
Since the x intercepts of each graph closest to each other are 2 and 3, the line must cut through 2.5. The coordinates for graphs are now (2.5,y1) and (2.5,y2). Substituting 2.5 into the equations gives the the y coordinates as 3.375 and -2.5. No need to use 2.5 to calculate distance since the gradient doesn't exist, so 3.374+2.5=5.875.

Not sure if this is right, but if it is could someone tell me how to work it out in a more calculus kind of way since this doesn't seem at all like calculus to me.
 
It's a nice attempt, but not exactly right. You can't make the assumption that since they're between 2 and 3 then it must be 2.5.
Take this graph, [tex]y=\frac{8x-16}{5-2x}[/tex]. It's x-intercept is 2 but it has an asymptote at 2.5 so this is an extreme example of how it doesn't work.

If we have y=f(x) and y=g(x) then we let them equal each other so we have f(x)=g(x) then y=f(x)-g(x), we are finding x where the graphs of f(x) and g(x) are equal, or in other words, where f(x)-g(x)=0. But since in our case they don't cross, we are trying to find x where they are closest to each other. We are trying to find where y is a minimum.
 
How do you find the x intercept?
 
Whats g(x)? Is it f(x) for the 2nd equation?
 
I do, but we don't use g(x), only f(x). Could you show me the first steps of working to get this? I'm pretty confused still.
 
Right, so we have the graph y=x3+x2-20x+34. This graph tells us the distance between our function f(x) and g(x) since we are finding the difference between them. So at x=2, it will give us y=6 so that means the distance at x=2 between f(x) and g(x) is 6. How do you find where the minimum distance is?
 
Yea but I'm having a mental block at the moment.
 
It's really simple! On the graph it tells us the distance between the two original curves f(x) and g(x). At x=2 we have y=6 which tells us the distance between the two curves is 6, at x=3 the distance between them is 10. Obviously the minimum distance between them is then the minimum point on the graph!
If I asked you to find the minimum of the parabola y=x2+2x using calculus, what would you do?
 
EDIT: Never mind with that, bit late
 
Don't think about finding what the actual distance is yet, we just want to find the x-value where the minimum occurs, which is where calculus comes into play. After we do this we'll substitute it back into equation to find the distance.
 
Is the minimum the turning point, so I need to find when the gradient equals zero?
 
You should be more confident in your abilities, rather:
"The minimum is the turning point, so I need to find when the gradient equals zero!"

:-p
 
f'(x)=3x^2+2x-20
I can't seem to manage factorizing this now, although I may be going in the wrong direction.
 
I use the 2.27 since the question asks when X is positive, then substitute X into f(x)-g(x) and that gives me 5.45 (2dp).
 
You should keep it in surd form. But we were told that x>0, so you can scrap the negative value. Now that you know where the minimum occurs, it's pretty obvious how close you are to the answer :smile:

Just as a note, cubics always cross the x-axis somewhere. Our difference function f(x)-g(x) crossed in the negative values of x, which means that f(x) equals to g(x) at some point (since their difference is 0) so if the question never restricted us to x>0 then the closest distance would be boring and just require us to solve the cubic f(x)-g(x)=0. This would be a harder question too...
 
I use the 2.27 since the question asks when X is positive, then substitute X into f(x)-g(x) and that gives me 5.45 (2dp).

Don't think you saw before you posted :P
 
Woo! Thank you so much for all your help! :D :D :D