Find the maximum and minimum values of $f (x) = {sin}^{4} x + {cos}^{4} x x \in [0, \frac{π}{2}]$

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Solution

${cos}^{2} (x) {+ sin}^{2} (x) = 1 s q u a r i n g o n b o t h s i d e s w e g e t$

${sin}^{4} x + {cos}^{4} x + 2 {sin}^{2} x {cos}^{2} x = 1$

$∴ f (x) = {sin}^{4} x + {cos}^{4} x = 1 - 2 {sin}^{2} x {cos}^{2} x$

$= 1 - \frac{{sin}^{2} 2 x}{2}$

${sin}^{2} 2 x r a n g e i s [0, 1]$

$0 \leq {sin}^{2} 2 x \leq 1$

$0 \geq - {sin}^{2} 2 x \geq - 1$

$0 \geq - \frac{{sin}^{2} 2 x}{2} \geq \frac{- 1}{2}$

$1 \geq 1 - \frac{{sin}^{2} 2 x}{2} \geq 1 - \frac{1}{2}$

$1 \geq {sin}^{4} x + {cos}^{4} x \geq \frac{1}{2}$

$∴$ minimum of $f (x)$ is $\frac{1}{2}$ and maximum of $f (x)$ is $1$

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