Name: Local Maxima and Minima of a Function
Uploaded: 2019-07-19
Duration: 14 min 17 s
Description: Do you know what is minima or maxima of a function? This video also explains how to find local maxima and minima by differentiation in a detailed manner.

Question

Find the minimum value of

f (x) = 8^{x} + 8^{- x} - 4 (4^{x} + 4^{- x}) \forall x \in R

Accepted Answer

Given, $f (x) = 8^{x} + 8^{- x} - 4 (4^{x} + 4^{- x}) \forall x \in R$

$\Rightarrow f^{'} (x) = 8^{x} \ln (8) - 8^{- x} \ln (8) - 4 (4^{x} \ln (4) - 4^{- x} \ln (4))$

$\Rightarrow f^{'} (x) = {(2^{x})}^{3} \ln (2^{3}) - {(\frac{1}{2^{x}})}^{3} \ln (2^{3}) - 4 ({(2^{x})}^{2} \ln (2^{2}) - {(\frac{1}{2^{x}})}^{2} \ln (2^{2}))$

$\Rightarrow f^{'} (x) = 3 ({(2^{x})}^{3} - {(\frac{1}{2^{x}})}^{3}) - 8 ({(2^{x})}^{2} - {(\frac{1}{2^{x}})}^{2}) \ln (2)$

$\Rightarrow f^{'} (x) = 3 (2^{x} - \frac{1}{2^{x}}) ({(2^{x})}^{2} + {(\frac{1}{2^{x}})}^{2} + 1) - 8 (2^{x} - \frac{1}{2^{x}}) (2^{x} + \frac{1}{2^{x}}) \ln (2)$

$\Rightarrow f^{'} (x) = (3 ({(2^{x})}^{2} + {(\frac{1}{2^{x}})}^{2} + 2 - 1) - 8 (2^{x} + \frac{1}{2^{x}})) (2^{x} - \frac{1}{2^{x}}) \ln (2)$

$\Rightarrow f^{'} (x) = (3 {(2^{x} + \frac{1}{2^{x}})}^{2} - 8 (2^{x} + \frac{1}{2^{x}}) - 3) \frac{(2^{x} - 1) (2^{x} + 1)}{2^{x}} \ln (2)$

$\Rightarrow f^{'} (x) = (3 (2^{x} + \frac{1}{2^{x}}) + 1) (2^{x} + \frac{1}{2^{x}} - 3) \frac{(2^{x} - 1) (2^{x} + 1)}{2^{x}} \ln (2)$

$\Rightarrow f^{'} (x) = (3 (2^{x} + \frac{1}{2^{x}}) + 1) ({(2^{x})}^{2} - 3 (2^{x}) + 1) \frac{(2^{x} - 1) (2^{x} + 1)}{2^{2 x}} \ln (2)$

$\Rightarrow f^{'} (x) = [(3 (2^{x} + \frac{1}{2^{x}}) + 1) \frac{(2^{x} + 1)}{2^{2 x}} \ln (2)] (2^{x} - \frac{3 - \sqrt{5}}{2}) (2^{x} - \frac{3 + \sqrt{5}}{2}) (2^{x} - 1)$

Here, $f^{'} (x)$ changes its sign from negative to positive about $x = 0, \log_{2} (\frac{3 + \sqrt{5}}{2})$ i.e. the given function has two points of local minima at $x = 0 & x = \log_{2} (\frac{3 + \sqrt{5}}{2})$ .

$\Rightarrow f (0) = 8^{0} + 8^{- 0} - 4 (4^{0} + 4^{- 0}) & f (\ln (\frac{3 + \sqrt{5}}{2})) = 8^{\log_{2} (\frac{3 + \sqrt{5}}{2})} + 8^{- \log_{2} (\frac{3 + \sqrt{5}}{2})} - 4 (4^{\log_{2} (\frac{3 + \sqrt{5}}{2})} + 4^{- \log_{2} (\frac{3 + \sqrt{5}}{2})})$

$\Rightarrow f (0) = - 6 & f (\ln (\frac{3 + \sqrt{5}}{2})) = {(\frac{3 + \sqrt{5}}{2})}^{3} + {(\frac{3 - \sqrt{5}}{2})}^{3} - 4 ({(\frac{3 + \sqrt{5}}{2})}^{2} + {(\frac{3 - \sqrt{5}}{2})}^{2})$

$\Rightarrow f (0) = - 6 & f (\ln (\frac{3 + \sqrt{5}}{2})) = (\frac{3 + \sqrt{5}}{2} + \frac{3 - \sqrt{5}}{2}) (\frac{9 + 5 + 6 \sqrt{5}}{4} + \frac{9 + 5 + 6 \sqrt{5}}{4} - 1) - 4 (\frac{9 + 5 + 6 \sqrt{5}}{4} + \frac{9 + 5 + 6 \sqrt{5}}{4})$

$\Rightarrow f (0) = - 6 & f (\ln (\frac{3 + \sqrt{5}}{2})) = - 10$

Hence, the minimum value of the given function is equal to $- 10$ .

Find the minimum value of $f (x) = 8^{x} + 8^{- x} - 4 (4^{x} + 4^{- x}) \forall x \in R$

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Important Questions on Application of Derivatives

Learn and Prepare for any exam you want !

Find the minimum value of $f (x) = 8^{x} + 8^{- x} - 4 (4^{x} + 4^{- x}) \forall x \in R$