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# Function functioned my brain

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, 8 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 $$f(79) = 25\cdot 79 = 1975$$ · 3 years, 8 months ago

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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, $$\text{f(79) = 1975}$$ · 3 years, 8 months ago

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How do you know that you can differentiate this function? Why must it even be continuous? Staff · 3 years, 8 months ago

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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 · 3 years, 8 months ago

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