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[SOLVED] s8.3.7.6 minimum vertical distance

karush

Well-known member
Jan 31, 2012
2,685
S8.3.7.6. What is the minimum vertical distance between the parabolas
$$y = x^2+1 \textit{and } y = x- x^2$$

Ok I think what question is ... The vertical distance between vertex's
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
658
S8.3.7.6. What is the minimum vertical distance between the parabolas
$$y = x^2+1 \textit{and } y = x- x^2$$

Ok I think what question is ... The vertical distance between vertex's

I interpret the question as the minimum of $|y_1 -y_2|$ where $y_1=x^2+1$ and $y_2=x-x^2$

... which would be a distance equal to 7/8
 
Last edited:

karush

Well-known member
Jan 31, 2012
2,685
$(x^2+1)'=2x$. Then 2x=0 so x=0
$(x-x^2)'=1-2x$ then 1-2x=0 so x=.5

(0)^2+1=1 And (.5)-(.5)^2=.25
Vertical distance is
$1-.25=.75$
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
658
$(x^2+1)'=2x$. Then 2x=0 so x=0
$(x-x^2)'=1-2x$ then 1-2x=0 so x=.5

(0)^2+1=1 And (.5)-(.5)^2=.25
Vertical distance is
$1-.25=.75$
You calculated the vertical difference between the respective y-values of each vertex ... only problem is, the two vertices are not vertically aligned. The vertex of $y=x^2+1$ is at $(0,1)$ and the vertex of $y=x-x^2$ is $\left(\frac{1}{2},\frac{1}{4}\right)$

To get the idea, try the same question with these two functions ...

What is the minimum vertical distance between $y = x^2+1$ and $y = x-1$

... obviously, $y=x-1$ doesn't have a vertex.
 

karush

Well-known member
Jan 31, 2012
2,685
Ok well this was from exercises in doing min/max problems where all the examples were solved by $f'(x)=0$

Yes the word distance was ackward as opposed to difference

This was an even problem # so no answer was given so you are probably correct
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
You are still ignoring skeeter's first response! No, this not asking for the distance between vertices, it is asking for the minimum of \(\displaystyle |y_1- y_2|= |x^2+ 1- x+ x^2|= |2x^2- x+ 1|\). That (ignoring the absolute value) is a parabola, \(\displaystyle 2(x^2- x/2+ 1/16- 1/16)+ 1= 2(x- 1/4)^2+ 15/16\). That's always positive so we can drop the absolute value. It has a minimum value of 15/16 when x= 1/4.

You could also use "f'(x)= 0" with f(x)= 2x^2-x+ 1. f'(x)= 4x- 1= 0. which gives x= 1/4 again.