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Differential Calculus

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

kavita Sonawane

14 Years agoGrade

1 Answer

Jitender Singh

12 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

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