I think it is an algebra problem (1)

Calculus Level 4

5 f ( x 2 ) + 3 f ( 1 x 2 ) = 2 x + 3 \large 5f(x^2)+3f\left(\frac{1}{x^2}\right)=2x+3

How many functions f : R + → R f: \mathbb{R}^+ \rightarrow \mathbb{R} are there that satisfy the above equation?


The answer is 1.

Aman Sharma by Aman Sharma , 25, India

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2 solutions

Pranjal Jain
Nov 1, 2014

5 f ( x 2 ) + 3 f ( 1 x 2 ) = 2 x + 3 5f(x^{2})+3f(\frac{1}{x^{2}})=2x+3

On substituting x → 1 x x\rightarrow\frac{1}{x} , 5 f ( 1 x 2 ) + 3 f ( x 2 ) = 2 x + 3 5f(\frac{1}{x^{2}})+3f(x^{2})=\frac{2}{x}+3

On eliminating f ( 1 x 2 ) f(\frac{1}{x^{2}}) from above two equations, we will get f ( x 2 ) = 3 8 + 5 x 8 − 3 8 x f(x^{2})=\frac{3}{8}+\frac{5x}{8}-\frac{3}{8x} ⇒ f 1 ( x ) = 3 8 + 5 x 8 − 3 8 x ; f 2 ( x ) = 3 8 − 5 x 8 + 3 8 x \Rightarrow f_{1}(x)=\frac{3}{8}+\frac{5\sqrt{x}}{8}-\frac{3}{8\sqrt{x}}; f_{2}(x)=\frac{3}{8}-\frac{5\sqrt{x}}{8}+\frac{3}{8\sqrt{x}}

f 1 ( x ) + f 2 ( x ) = 3 4 f_{1}(x)+f_{2}(x)=\frac{3}{4}

∫ 0 4 3 4 d x = 3 4 × 4 = 3 \displaystyle\int_{0}^{4} \frac{3}{4} dx=\frac{3}{4}×4=\boxed{3}

8 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

As pointed out in the comments, f ( x 2 ) = x f(x^2 ) = x does not imply that f ( x ) = − x f(x) = - \sqrt{x} . For example, the first expression gives us f ( 1 ) = 1 f(1) = 1 and the second gives us f ( 1 ) = − 1 f(1) = -1 .

f 2 f_{2} is no solution because one gets − 2 ⋅ x + 3 -2\cdot x+3 for the expression above!!!

Andreas Wendler - 5 years ago

I did it the same way too.

Aakarshit Uppal - 6 years, 4 months ago

5 f ( x 2 ) + 3 f ( 1 x 2 ) = 2 x + 3 5f(x^2) + 3f(\frac{1}{x^2}) = 2x + 3

Substitute x → 1 x x \to \frac{1}{x}

5 f ( 1 x 2 ) + 3 f ( x 2 ) = 2 x + 3 5f(\frac{1}{x^2}) + 3f(x^2) = \dfrac{2}{x} + 3

On solving these two equations and eliminating f ( 1 x 2 ) f(\frac{1}{x^2}) , we get

8 f ( x 2 ) = 3 + 5 x − 3 x 8f(x^2) = 3 + 5x - \dfrac{3}{x}

8 f ( x ) = 3 + 5 x − 3 x 8f(x) = 3 + 5\sqrt{x} - \dfrac{3}{\sqrt{x}}

8 x f ( x ) = 3 x + 5 x − 3 8\sqrt{x} f(x) = 3\sqrt{x} + 5x - 3

( x ) 2 ( 8 f ( x ) − 3 ) 2 = ( 5 x − 3 ) 2 (\sqrt{x})^2(8f(x) - 3)^2 = (5x-3)^2

64 x f 2 ( x ) + 9 x − 48 x f ( x ) = 25 x 2 + 9 − 30 x 64xf^2(x) + 9x - 48xf(x) = 25x^2 + 9 - 30x

Let f 1 ( x ) and f 2 ( x ) f_1(x) \text{ and } f_2(x) be 2 solutions.

f 1 ( x ) + f 2 ( x ) = 48 x 64 x = 3 4 f_1(x) + f_2(x) = \dfrac{48x}{64x} = \dfrac{3}{4}

The integral is then equal to 3 \boxed{3} .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

As pointed out in the comments, f ( x 2 ) = x f(x^2 ) = x does not imply that f ( x ) = − x f(x) = - \sqrt{x} . For example, the first expression gives us f ( 1 ) = 1 f(1) = 1 and the second gives us f ( 1 ) = − 1 f(1) = -1 .

First one can not artificially induce a second solution by simply square an equation! Indeed on my opinion there exists only one solution for f(x) correctly determined in the first part of your posting.

Furthermore you went wrong since you may not handle your equation like a polynomial (via Vieta) because 'x' is contained!

Therefore the problem seems to be irrelevant!

Andreas Wendler - 5 years ago

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