# 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

Jitender Singh IIT Delhi
9 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