Lagrange's mean value theorem

Calculus Level 5

Function f ( x ) f(x) is such that f ( x ) = a ln ⁡ x + x 2 2 f(x)=a\ln \ x +\dfrac{x^2} 2 , where a > 0 a > 0 is a parameter.

If f ( x 1 ) − f ( x 2 ) x 1 − x 2 > 2 \dfrac{f(x_1)-f(x_2)}{x_1-x_2}>2 for all x 1 , x 2 ∈ ( 0 , + ∞ ) x_1, x_2 \in (0,+\infty) and x 1 ≠ x 2 x_1 \ne x_2 , what is the range of a a ?

( 0 , 1 ) (0,1) ( 0 , 1 ] (0,1] ( 1 , + ∞ ) (1,+\infty) [ 1 , + ∞ ) [1,+\infty)

Alice Smith by Alice Smith , 20, China

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

This is a solution. It is not a proof.

I note that the case of x 1 = x 2 x_1=x_2 is precluded. That is required to prevent a division by 0 0 .That permits the closed lower limit as the function is not defined there. I note that if a < 1 a<1 than the function reaches values less than 2 2 . Values of a a such that the function has values ≥ 2 \geq 2 are required. Of the four answers provided, [ 1 , ∞ ) [1,\infty) best answers the question.

0 Helpful 0 Interesting 0 Brilliant 1 Confused

Let g ( x ) = f ( x ) − 2 x . g(x)=f(x)-2x. We can see that f ( x 1 ) − f ( x 2 ) x 1 − x 2 > 2 \frac{f(x_1)-f(x_2)}{x_1-x_2}>2 for all positive values of x 1 x_1 and x 2 x_2 if and only if g ( x 1 ) − g ( x 2 ) x 1 − x 2 > 0 \frac{g(x_1)-g(x_2)}{x_1-x_2}>0 for all positive values of x 1 x_1 and x 2 . x_2. Now we are going to prove that the latter condition is true if and only if a ≥ 1. a\geq 1. We are going to prove first the implication:

if g ( x 1 ) − g ( x 2 ) x 1 − x 2 > 0 \frac{g(x_1)-g(x_2)}{x_1-x_2}>0 for all positive values of x 1 x_1 and x 2 , x_2, then a ≥ 1. a\geq 1.

Indeed, if g ( x 1 ) − g ( x 2 ) x 1 − x 2 > 0 \frac{g(x_1)-g(x_2)}{x_1-x_2}>0 for all positive values of x 1 x_1 and x 2 , x_2, then g ′ ( x ) ≥ 0 g'(x)\geq 0 for all x . x. Then a x + x − 2 = ( x − 1 ) 2 + a − 1 x ≥ 0 \frac{a}{x}+x-2=\frac{(x-1)^2+a-1}{x}\geq 0 for any positive x . x. Then making x = 1 x=1 in the previous inequality we obtain that a − 1 ≥ 0 a-1\geq 0 and, therefore, a ≥ 1 a\geq 1

Let us prove now the converse implication:

if a ≥ 1 , a\geq 1, then g ( x 1 ) − g ( x 2 ) x 1 − x 2 > 0 \frac{g(x_1)-g(x_2)}{x_1-x_2}>0 for all positive values of x 1 x_1 and x 2 . x_2.

Using that g ′ ( x ) = ( x − 1 ) 2 + a − 1 x , g'(x)=\frac{(x-1)^2+a-1}{x}, as we saw before, we can see that when a ≥ 1 , a\geq 1, then g ′ ( x ) > 0 g'(x)>0 for all x x in the interval ( 0 , 1 ) (0, 1) and again g ′ ( x ) > 0 g'(x)>0 for all x x in the interval ( 1 , ∞ ) . (1, \infty). Then, regardless the value of the function g ′ g' at 1 (that is greater than or equal to zero) and due to the fact that g ( x ) g(x) is continuous on the interval ( 0 , ∞ ) (0, \infty) , we obtain that g ( x ) g(x) is strictly increasing on the interval ( 0 , ∞ ) . (0, \infty). Therefore, g ( x 1 ) − g ( x 2 ) x 1 − x 2 > 0. \frac{g(x_1)-g(x_2)}{x_1-x_2}>0. for all positive values of x 1 x_1 and x 2 . x_2.

Then according to what we have proved above, we have that f ( x 1 ) − f ( x 2 ) x 1 − x 2 > 2 \frac{f(x_1)-f(x_2)}{x_1-x_2}>2 for all positive values of x 1 x_1 and x 2 x_2 if and only if a ≥ 1. a\geq 1. Therefore, the range of a a is the interval [ 1 , ∞ ) . \boxed{[1, \infty)}.

Arturo Presa - 2 years, 3 months ago

2 pending reports

Vote up reports you agree with

×

Problem Loading...

Note Loading...

Set Loading...