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Screenshot_1.png Let f be a polynomial function such that f(3x)=f'(x)f"(x) for x belongs to R Then

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@Priti

If f(x) is of first degree its second derivative is identically null, so also f(x) would have to be identically null. to satisfy the equation f(3x)=f'(x)f''(x)

Let then f(x) be a generic polynomial of degree n≥2. Then f'(x) will have degree (n−1) and f''(x) degree (n−2)

Now, the product f'(x)⋅f''(x) is a polynomial of degree (n−1)+(n−2) and as two polynomials can be equal for every x only if they have the same degree:

\\n=(n-1)+(n-2)\\n=3\\

that is f(x) must be of third degree

\\f(x)=ax^3+bx^2+cx+d\\f'(x)=3ax^2+2bx+c\\f''(x)=6ax+2b\\

\\now,\\given,f(3x)=f'(x)f''(x)\\replace,\:\:x\rightarrow 3x\\27ax^3+9bx^2+3cx+d = (3ax^2+2bx+c)(6ax+2b)

Equating the coefficients of the same degree we get:

\\27a = 18a^2\\a=3/2\\similarly,b,c,d\:become\:0\\

The polynomial which satisfies the equation is then

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

 

f (x) = a.r3 given, f(3x) = f'(x) replace, x —+ 3x 270 = = 3/2 similarly, b, c, d become 0

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himanshu.meshram

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