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

$\frac{7}{8}$

01:31

Wen Z.

01:06

Amrita B.

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 7

Optimization Problems

Derivatives

Differentiation

Volume

Missouri State University

Campbell University

Idaho State University

Boston College

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we're asked to find the minimum vertical distance between the parade. Bolas Y equals X squared plus one and y equals X minus X squared. Okay, big bar. Well, to do this first nous the vertical distance between these two dfx. This is the absolute value of X squared plus one minus X minus X squared, which is the absolute value of two X squared minus X plus one. You know, it's like now we know that the distance G is well minimized. Black when it's square d squared, which I'll call f is minimized as well. Now the function f of X, this is D squared of X, which is no, it's a piece of shit. I could have two x squared minus X plus one squared Mhm. All right. Now, in order to find the minimum vertical the minimum value of f obey dot com I'm going to take the derivative of F. So we have F Private X is by the chain rule two times two x squared minus X plus one times four x minus one So here is equal to zero. So the critical values satisfy two X squared minus X plus one equals zero or four X minus one equals zero. Now the discriminate of this quadratic equation Delta, this is B Square, which is one minus four times a, which is two times See, So this is one minus eight, which is negative seven, which is less than zero. So this first equation has no real solutions. The second equation has the real solution. X equals 1/4. Now, at this point, we're only considering Nick convinced himself. We haven't restrained our function to an interval, however, noticed that this our function f of X is in fact approaches positive infinity as X approaches plus or minus infinity. Therefore, it follows that F has an absolute minimum on its domain, and it's absolute minimum must occur at one of the critical values. So it follows that F has absolute quite sure minimum at X equals 1/4 and therefore, yeah, the minimum vertical distance is D of 1/4 which is the absolute value, and I'm of two times when fourth squared is 1/16 minus 1/4 plus one. This is the same as 1/8 minus 1/4 plus one. This is 1/8 minus 2/8 plus 8/8 which is nine minus two is 7/8 absolute value of 7/8 which is just 7/8. What? This is our answer.

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