smallest no.

For x ≥ 0, the smallest value of the function f(x) = 4x2+8x +13 6(x+1) is :

4 Answers

Aditya Agarwal ·


man111 singh ·

\hspace{-16}$Let $\bf{y=\frac{4x^2+8x+13}{6(x+1)}\;,}$ where $\bf{x\geq 0}$\\\\\\ Let $\bf{(x+1)=t.}$ here $\bf{(x+1)\geq 1}$ means $\bf{t\geq 1}$\\\\\\ So $\bf{y=\frac{4(t-1)^2+8(t-1)+13}{6t}=\frac{4t^2+9}{6t}}$\\\\\\ $\bf{6y=\frac{4t^2+9}{t} = \left(4t+\frac{9}{t}\right)\;,}$ where $\bf{t\geq 1}$\\\\\\ Now using $\bf{A.M\geq G.M\;,}$ we get\\\\\\ $\bf{\frac{4t+\frac{9}{t}}{2}\geq \sqrt{4t\cdot \frac{9}{t}}=6}$\\\\\\ So $\bf{\left(4t+\frac{9}{t}\right)\geq 6 \times 2}$\\\\\\ So $\bf{6y\geq 6\times 2\Rightarrow y\geq 2}$\\\\\\ and equality hold when $\bf{4t=\frac{9}{t}\Rightarrow t = \frac{3}{2}}$\\\\\\ So $\bf{y_{Min.} = 2}$ when $\bf{x+1=\frac{3}{2}\Rightarrow x = \frac{1}{2}}$.

Himanshu Giria ·

is :

Aditya Agarwal ·

You could even do it in this way:-

f(x) = 4(x+1)2 + 96(x+1)
→ 23(x+1) + 32(x+1)

now find x for which f'(x)=0 and put it in f(x) to get min value of f(x).
To be sure that it IS the min. value and not the max. find f"(x) for the value of x you found from f'(x) and see whether its <0.

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