Maxima and Minima — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsMaxima and Minima4 Marks
Question
Find the maximum and minimum values of the function $\text{f}(\text{x})=\frac{4}{\text{x}+2}+\text{x}$
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Answer
Given, $\text{f}(\text{x})=\frac{4}{\text{x}+2}+\text{x}$ $\Rightarrow\text{f}'(\text{x})=-\frac{4}{(\text{x}+2)^{2}}+1$ For a local maxima or a local minima, We must have f'(x) = 0 $\Rightarrow-\frac{4}{(\text{x}+2)^{2}}+1=0$ $\Rightarrow-\frac{4}{(\text{x}+2)^{2}}=-1$ $(\text{x}+2)^{2}=\pm2$ $\Rightarrow \text{x}=0 \ \text{and} -4$ Thus, x = 0 and x = -4 are the possible of local maxima or local minima. Now, $\text{f}'(\text{x})=\frac{8}{(\text{x}+2)^{3}}$ At x = 0 $\text{f}''(0)=\frac{8}{(2)^{3}}=1>0$ So, x = 0 is a point of local minimum. The local minimum value is given by $\text{f}(0)=\frac{4}{(0+2)}+0=2$
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