If f(3x) = f(3) + f(x) , then prove that : f(1) = f(3) =f(9) = f(27) = f(81) = 0

[Riwaj]

Q.1) If f(3x) = f(3) + f(x) and x\in R , then prove that : f(1) = f(3) =f(9) = f(27) = f(81) = 0 .
Q.2) If g(x) = x² -4x + 3 and g(x+1) = g(x-1) , then find the value of x .

 

[Riwaj]

(Sweating) its very urgent

 

[MarkFL]

Q.1) If f(3x) = f(3) + f(x) and x\in R , then prove that : f(1) = f(3) =f(9) = f(27) = f(81) = 0 .

Okay, we are told:

\(\displaystyle f(3x)=f(3)+f(x)\)

Suppose we let $x=3^a$, where $a\in\mathbb{N_0}$...we then get:

\(\displaystyle f\left(3^{a+1}\right)=f(3)+f\left(3^a\right)=f(3)+\left(f(3)+f\left(3^{a-1}\right)\right)=?\)

 

[Riwaj]

isn't there any simper way .

 

[HallsofIvy]

Did you understand what MarkFL was telling you? Since f(3x)= f(3)+ f(x), taking x= 1, f(3)= f(3(1))= f(3)+ f(1) so f(1)= ?
f(9)= f(3(3))= f(3)+ f(3)= 2f(3).
f(27)= f(3(9))= f(3)+ f(9)= f(3)+ 2f(3)= 3f(3).
f(81)= f(3(27))= f(3)+ f(27)= f(3)+ 3f(3)= 81f(3).

Q.2) If g(x) = x² -4x + 3 and g(x+1) = g(x-1) , then find the value of x .

g(x+ 1)= (x+ 1)²- 4(x+ 1)+ 3.
g(x- 1)= (x- 1)²- 4(x- 1)+ 3

Multiply those out and set them equal. You have a linear equation to solve for x.

 

[MarkFL]

Okay, we are told:

\(\displaystyle f(3x)=f(3)+f(x)\)

Suppose we let $x=3^a$, where $a\in\mathbb{N_0}$...we then get:

\(\displaystyle f\left(3^{a+1}\right)=f(3)+f\left(3^a\right)=f(3)+\left(f(3)+f\left(3^{a-1}\right)\right)=?\)

If carry this out, we find that:

\(\displaystyle f\left(3^a\right)=a\cdot f(3)\)

This, along with the given equation should suggest that $f$ could be a logarithmic function. Can you find a logarithmic function that satisfies the given equation, but for which

\(\displaystyle f(1)=f(3)=f(9)=f(27)=f(81)=0\)

is not true?

 

[MarkFL]

Q.2) If g(x) = x² -4x + 3 and g(x+1) = g(x-1) , then find the value of x .

Another approach to this problem, as we saw in a previously posted problem of yours, is to use the axis of symmetry. Since:

\(\displaystyle g(x+1)=g(x-1)\) then the axis of symmetry $x_S$ has to be:

\(\displaystyle x_S=\frac{(x+1)+(x-1)}{2}=x\)

In fact, we must have:

\(\displaystyle g(x+k)=g(x-k)\) where $k\in\mathbb{R}$

But, we also know:

\(\displaystyle x_S=-\frac{b}{2a}\)

For the general quadratic $ax^2+bx+c$...so equate the two expressions for the axis of symmetry and solve for $x$. What do you get?