Question

f, f' both are continuous and f '' exist on [a, b]. Prove that if f has at least three distinct zeros in [a, b], then the equations f(x) + f ''(x) = 2f ' (x) has at least one root in [a, b]​

Answer 1

The value of f₁ ( x + y ) + f₁( x - y ) equals 2f₁ ( x ) f₁ ( y )

Given f ( x ) = aˣ ( a > 0 )

• f ( x ) = f₁ ( x ) + f₂ ( x ) , where f₁ ( x ) is an even function and f₂(x) is an odd function.

We can take

• f1 ( x ) = ( aˣ + a^{-x} ) / 2 since f1 is even

and

• f2 ( x ) = ( aˣ - a^{-x} ) / 2 since f2 is odd
• f₁( x + y ) + f₁ ( x - y ) = ( a^{x + y} + a^{- (x+y)}) / 2 + ( a^{x-y} + a^{-(x -y)} ) / 2
• f₁ ( x + y ) + f₁ ( x - y ) = ( a^{x} + a^{-x} ) ( a^{y} + a^{-y} ) / 2

Hence,

• f₁( x + y ) + f₁( x - y ) = 2 f1 ( x ) . f1 ( y )

Answer 2

$\large{\underline{\underline{\mathsf\blue{Required\:Answer}}}}$

• f1(x) = (a + a^{-x}) / 2 since f1 is even

• f2 (x) = (a - a^{-x})/2 since f2 is odd

→ f₁(x + y) + f₁(x - y) = (a^{x + y} + a^{ (x+y)])/2 + (a^{x-y} + a^{-(x -y)} ) / 2 • f₁ (x + y) + f₁ (x - y) = (a^{x} +a^{-x}) (a^{y} + a^{-y} ) / 2

Hence,

• . f₁(x+y)+f₁(x - y) = 2 f1 (x). f1(y)

Sorry!