Question

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Answer 1

$f(2x+1) = x^{4} +4x^{2}$

2x + 1 = t :

x = (t-1)/2

$f(t) = (\frac{t-1}{2} )^{4} +4(\frac{t-1}{2} )^{2}=\frac{(t-1)^{4}}{16} +4\frac{(t-1)^{2}}{4} =\frac{(t-1)^{4}}{16}+(t-1)^{2}$

Значит :

$f(x) } =\frac{(x-1)^{4}}{16}+(x-1)^{2}$

$f'(x) = \frac{4(x-1)^{3} }{16} +2(x-1)=\frac{(x-1)^{3} }{4} +2(x-1)$

$f'(3)=\frac{(3-1)^{3} }{4} +2(3-1)=\frac{2^{3} }{4} +2*2 = \frac{8}{4} +4 = 2 +4 = 6$

Ответ : 6