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I was working with this function problem but, I got stuck. Here's the problem: Let f(x) be a function with the two properties_ a. For any two real numbers x and y, f(xy)=x.f(y) and _b. f(1)=25. What is the value of f(79)?

Note by Towkir Hossain 3 years, 10 months ago

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Given \(f(xy) = x\cdot f(y)\) and \(f(1) = 25\),

Put \(y = 1\), we get \(f(x) = x\cdot f(1) = 25x\Rightarrow f(x) = 25x\)

So Put \(x = 79\), we get \(f(79) = 25\cdot 79 = 1975\)

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Hi, this seems simple:

\(\text{f(xy) = xf(y)}\)

Derivate with respect to \(y\) taking \(x\) constant to get:

\(\text{xf'(xy) = xf'(y)}\)

\(\Rightarrow \text{f'(xy) = f'(x) = k}\) (say) Integrate to get \(\text{f(x) = kx + c}\)

\(\text{f(xy) = x f(y)} \Rightarrow \text{kxy + c = kxy + xc} \) \(\forall\) \(x \in R\)

Hence, \(\text{c=0}\).

Now, \(\text{f(x) = kx}\), and \(f(1) = 25 \Rightarrow \text{f(x) = 25x}\)

Hence, \(\text{f(79) = 1975}\)

How do you know that you can differentiate this function? Why must it even be continuous?

f(xy)=x.f(y) given now put y=1 f(x1)=xf(1) =25x (as f(1)=25) so now f(79)=25*79 =1975

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## Comments

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TopNewestGiven \(f(xy) = x\cdot f(y)\) and \(f(1) = 25\),

Put \(y = 1\), we get \(f(x) = x\cdot f(1) = 25x\Rightarrow f(x) = 25x\)

So Put \(x = 79\), we get \(f(79) = 25\cdot 79 = 1975\)

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Hi, this seems simple:

\(\text{f(xy) = xf(y)}\)

Derivate with respect to \(y\) taking \(x\) constant to get:

\(\text{xf'(xy) = xf'(y)}\)

\(\Rightarrow \text{f'(xy) = f'(x) = k}\) (say) Integrate to get \(\text{f(x) = kx + c}\)

\(\text{f(xy) = x f(y)} \Rightarrow \text{kxy + c = kxy + xc} \) \(\forall\) \(x \in R\)

Hence, \(\text{c=0}\).

Now, \(\text{f(x) = kx}\), and \(f(1) = 25 \Rightarrow \text{f(x) = 25x}\)

Hence, \(\text{f(79) = 1975}\)

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How do you know that you can differentiate this function? Why must it even be continuous?

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f(xy)=x.f(y) given now put y=1 f(x

1)=xf(1) =25x (as f(1)=25) so now f(79)=25*79 =1975Log in to reply