5352
- Aditya Agarwal Take f(x)= m and solve for D >= 0, where D is the discriminant of the new quadratic equation so formed.Upvote·0· Reply ·2014-04-13 22:29:25
Himanshu Giria thanx...
For x ≥ 0, the smallest value of the function f(x) = 4x2+8x +13 6(x+1) is :
-
UP 0 DOWN 0 0 4
4 Answers
535
1708\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 $\bf{x+1=\frac{3}{2}\Rightarrow x = \frac{1}{2}}$.
535You 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.
Akshay Ginodia i dnt think we need to find f'(x) because it is of the form a+1/a so its minimum value is 2- Aditya Agarwal Oh yes! Didn't occur to me. :)