What is the extreme (16a^4+16a)/(a^2+1-a) ?

The extreme (16a^4+16a)/(a^2+1-a) is Minimum(-1/2 ,-4)

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extreme (16a^4+16a)/(a^2+1-a)

extreme points

\frac{16 a ^{4} + 16 a}{a ^{2} + 1 - a}

Minimum (- \frac{1}{2}, - 4)

Solution steps

f^{'} (a) = 16 (2 a + 1)

Find intervals:

Decreasing

: - \infty < a < - \frac{1}{2},

Increasing

: - \frac{1}{2} < a < \infty

Plug

a = - \frac{1}{2}

into

\frac{16 a ^{4} + 16 a}{a ^{2} + 1 - a} : - 4

Minimum (- \frac{1}{2}, - 4)

f (x) = - 2 x^{2} - 5 x y - 6 y^{2} + 32 x + 63 y + 9

f (x) = 4 sin (x) - 3 cos (x)

y = (\frac{I x + 2 I - 1}{2 x + 3}) e^{2 x}

x + \frac{3}{2} x^{\frac{2}{3}}

+ ∣ x^{2} - 3 x + 2 ∣

What is the extreme (16a^4+16a)/(a^2+1-a) ?
The extreme (16a^4+16a)/(a^2+1-a) is Minimum(-1/2 ,-4)