4x-4-2x-2-17-max-

Question Number 146253 by mathdanisur last updated on 12/Jul/21

{(4 x^{4} - 2 x^{2} + 17)}_{m a x} = ?

Answered by iloveisrael last updated on 12/Jul/21

f {(x)}_{\min} = 4 {(\frac{1}{4})}^{2} - 2 (\frac{1}{4}) + 17

f {(x)}_{\min} = \frac{1}{4} - \frac{2}{4} + \frac{68}{4} = \frac{67}{4}, when x^{2} = \frac{1}{4}

f {(x)}_{\max} = \infty

Commented by mathdanisur last updated on 12/Jul/21

t h a n k y o u S e r

Answered by gsk2684 last updated on 12/Jul/21

4 (x^{4} - \frac{1}{2} x^{2}) + 17

4 ({(x^{2} - \frac{1}{4})}^{2} - \frac{1}{16}) + 17

4 {(x^{2} - \frac{1}{4})}^{2} - \frac{1}{4} + 17

4 {(x^{2} - \frac{1}{4})}^{2} + \frac{67}{4}

m i n i m u m i s \frac{67}{4}

m a x i m u m c a n n o t d e f i n e o n R

Commented by mathdanisur last updated on 12/Jul/21

t h a n k y o u S e r