What is the extreme f(x)=tsqrt(4-t) ?

The extreme f(x)=tsqrt(4-t) is Maximum(8/3 ,(16)/(3sqrt(3))),Minimum(4,0)

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extreme f(x)=tsqrt(4-t)

extreme points

f (x) = t 4 - t

Maximum (\frac{8}{3}, \frac{16}{3 3}), Minimum (4, 0)

Solution steps

f^{'} (t) = \frac{- 3 t + 8}{2 4 - t}

Find intervals:

Increasing

: - \infty < t < \frac{8}{3},

Decreasing

: \frac{8}{3} < t < 4

Plug

t = \frac{8}{3}

into

t 4 - t : \frac{16}{3 3}

Maximum (\frac{8}{3}, \frac{16}{3 3})

Plug

t = 4

into

t 4 - t : 0

Minimum (4, 0)

Maximum (\frac{8}{3}, \frac{16}{3 3}), Minimum (4, 0)

x^{2} + \frac{1}{x}

f (x) = x^{3} - x^{2} - x + 8, - 1 \leq x \leq 0

f (x) = \frac{12 x}{9 + x ^{2}}

f (x, y) = 4 - 2 x + 4 y - x^{2} - 4 y^{2}

f (x) = 2 x^{\frac{5}{3}} - 5 x^{\frac{4}{3}}

What is the extreme f(x)=tsqrt(4-t) ?
The extreme f(x)=tsqrt(4-t) is Maximum(8/3 ,(16)/(3sqrt(3))),Minimum(4,0)