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Find the polynomial function of degree 6 satifying limit xtends to zero [1+(f(x)/x^3]^1/x=e^2and has local maxima at x=1 and local minima at x=0,2

Find the polynomial function of degree 6 satifying limit xtends to zero [1+(f(x)/x^3]^1/x=e^2and has local maxima at x=1 and local minima at x=0,2

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1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
8 years ago
Ans:
\lim_{x\rightarrow o}(1+\frac{f(x)}{x^{3}})^{\frac{1}{x}} = e^{2}
\lim_{x\rightarrow o}(1+\frac{f(x)}{x^{3}})^{({\frac{x^{3}}{f(x)}}.\frac{f(x)}{x^{4}})} = e^{2}
e^{\lim_{x\rightarrow 0}\frac{f(x)}{x^{4}}} = e^{2}
{\lim_{x\rightarrow 0}\frac{f(x)}{x^{4}}} = 2
Also, we have
{\lim_{x\rightarrow 0}\frac{f(x)}{x^{3}}} = 0
So, it is clear from this this expression that f(x) has a degree greater than 3.
Let
f(x) = ax^{6}+bx^{5}+cx^{4}
{\lim_{x\rightarrow 0}\frac{ax^{6}+b^{5}+cx^{4}}{x^{4}}} = 2
\Rightarrow c= 2
f^{'}(x) = 5ax^{5}+5bx^{4}+4cx^{3}
f^{'}(1) = 6a(1)^{5}+5b(1)^{4}+4.2(1)^{3} = 0
6a+5b+8=0….....(1)
f^{'}(2) = 6a(2)^{5}+5b(2)^{4}+4.2(2)^{3} = 0
192a + 80b+64=0…......(2)
(1) \times 16
96a + 80b+128=0….....(3)
(2) – (3)
96a - 64 = 0
a = \frac{2}{3}
Put in (1)
b = -\frac{12}{5}
f(x)= \frac{2}{3}x^{6}-\frac{12}{5}x^{5}+2x^{4}
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

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