Can anyone help me with this thinking math question?

Let

x = number

S = sum of the number and its reciprocal

Therefore,

S = x + (1/x)

Differentiating the above,

dS/dx = 1 - 1/x^2

and setting dS/dx = 0,

dS/dx = 0 = 1 - 1/x^2

and solving for "x"

x^2 = 1 and

x = 1 and x = -1

and since the problem asks for a positive number, then

x= 1

is the answer.

CHECK:

Take the second derivative of the original equation for "S", i.e.,

d^2S/dx^2 = 1/x^3

and since

d^2S/dx^2 > 0. then the calculated value of x = 1 is the value of "x" that will satisfy the condition of the problem.

Hope this helps.

Let x be the number. Sum of the number and its reciprocal = x + 1/x. To find the minimum value of this, differentiate by x and let it equal 0:

(x + (1/x))' = 0

1 - 1/x^2 = 0

1 = 1/x^2

x^2 = 1

x = (+-)1

But x > 0 so x = 1.

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If you don't know calculus, there is another way... you could notice that f(1) = 2; and that for any x > 1, f(x) > 2.

But we also have that

f(x) = x + 1/x = 1/(1/x) + (1/x) = f(1/x);

So that if f(x) > 2 for all x > 1, then [replacing x with 1/x]

f(1/x) > 2 for all 1/x > 1, and [since f(1/x) = f(x)], we have

f(x) > 2 for all 1/x > 1, i.e.

f(x) > 2 for all x < 1.

So finally we have that f(x) > 2 for all x > 1 and for all x < 1; therefore x = 1 is the minimum point, for which f(1) = 2.

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Finally, you could use quadratic equations: noticing that f(1) = 2, we try and find an x which gives a lower value for f than 2; i.e. we need to find x such that

f(x) = x + 1/x < 2.

x^2 + 1 < 2x

x^2 - 2x + 1 < 0

(x + 1)^2 < 0.

There are clearly no solutions to this, since the LHS is manifestly positive and the RHS is manifestly negative. No x will satisfy this equation. Hence there are no x values which will make f(x) lower than 2; so the minimum is at f(1) = 2.

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I suppose you could also solve it with a perturbation method...

Say the value of x which minimises f(x) is called x*. Then consider a 'nearby' solution x* + Δx, where Δx is some number (doesn't need to be infinitesimal: consider it of order 0.1, and can be positive or negative). In that case, the minimum value of f(x) is given by

f(x*) = x* + 1/x*; and the value of f at the nearby solution is

f(x* + Δx) = x* + Δx + 1/(x* + Δx)

= [(x* + Δx)^2 + 1]/(x* + Δx)

= [x*^2 + 2x*Δx + Δx^2 + 1]/[x* + Δx].

In this case, we know that x* = 1 and f(x*) = 2, so that

f(1 + Δx) = [1 + 2Δx + Δx^2 + 1]/[1 + Δx]

= [2(Δx + 1)]/[1 + Δx] + Δx^2/[1 + Δx]

= 2 + Δx^2/[1 + Δx].

Now the second term on the RHS is definitely positive, so

f(1 + Δx) > f(1); ie moving x away from 1 a small amount in either direction increases f(x); so x = 1 must be its minimum.

You can solve this question easily if you are familiar with calculus

let f(x) = x + 1/x (number plus it's eciprocal)

now you can find the derivative of f(x)

d(f(x))/d(x) = 1 -1/(x^2), for getting the minimum value of function f(x), it's derivative should be equal to zero.

So, 1 - 1/(x^2) = 0

=> 1 = 1/(x^2)

=> x^2 = 1

=> x = +1 or -1, so the function is minimum at x = 1 if you don't consider negative numbers, f(1) = 2

If you aren't familiar with calculus, then the only way to do it is through hit and trial, I don't think there is any logic to prove it, other than you calculate it for some numbers and see the trend, that as the number x increases form 1, so does the function f(x).

Also you could explain it likwise -

f(1) = 2

now f(2) = 2 + 1/2 which is apparently greater than 2

so every f(x) is greater than x, as it is equalto x + 1/x > x

this is the only logical explanantion I see, hence as f(x) is always greater than x, therefore, the smaller the number the smaller f(x) is.

S = x + (1 / x)

the number would have to be 1. If you were to do this with any other number, the result would have to be greater.

The answer is 1

1 + 1/1 = 2

2 + 1/2 = 2 1/2 (this is bigger so that's not it)

3 + 1/3 = 3 1/3 (this is bigger so that's not it)

etc.

The larger the sum,the smaller the reciprocal

.

Find the minimum using calculus:

.

Stationary points: Solve subject to the restriction x > 0.

I get . Test nature ....

The minimum value occurs at x = 1 and the minimum value is S = 2.

f(n) = n + 1/n

= n + nˉ¹

f'(n) = 1 - nˉ²

= 1 - 1/(n²)

When f(n) is at a minimum, f'(n) = 0

1 - 1/(n²) = 0

n² - 1 = 0

(n+1)(n-1) = 0

n = -1, 1

Since n is positive, n=1

http://www.flickr.com/photos/dwread/3418441592/

f(x) = x + 1/x

f '(x) = 1 - 1/x^2 = 0

x = 1 , -1

at x = 1 there is a max pt

isnt it 1....?