Let f : ℝ $\to$ ℝ be a polynomial function of degree four having extreme values at x = 4 and x = 5. If $ \lim\limits_{x \to 0} \frac{f(x)}{x^2} = 5 $, then f(2) is equal to :
Solution
<p>$$\begin{aligned}
& \lim _{x \rightarrow 0} \frac{f(x)}{x^2}=5 \\
& \lim _{x \rightarrow 0} \frac{\left.a x^4+b x^3+c x^2+d x+e\right)}{x^2}=5 \\
& c=5 \text { and } d=e=0 \\
& f(x)=a x^4+b x^3+5 x^2 \\
& f^{\prime}(x)=4 a x^3+3 b x^2+10 x \\
& =x\left(4 a x^2+3 b x+10\right)
\end{aligned}$$</p>
<p>has extremes at 4 and so $f^{\prime}(4)=0 \& f^{\prime}(5)=0$</p>
<p>so $\mathrm{a}=\frac{1}{8} \& \mathrm{~b}=\frac{-3}{2}$</p>
<p>so $f(2)=\frac{1}{8} \times 2^4-\frac{3}{2} \times 2^3+5 \times 2^2$</p>
<p>$=2-12+20=10$</p>
About this question
Subject: Mathematics · Chapter: Application of Derivatives · Topic: Tangents and Normals
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