Function given: $$f(\alpha, b,\delta) = \left(a_0 + a_2\cdot(\alpha-\delta)^2 + a_4\cdot(\alpha-\delta)^4\right)\cdot\left(1-\frac{b-1}{2b}\right)\cdot e^{j\cdot \pi} + \left(a_0 + a_2\cdot(\alpha+\delta)^2 + a_4\cdot(\alpha+\delta)^4\right)\cdot\frac{b-1}{2b}$$
Finding the derivative of a function: $$\frac{d f(\alpha, b,\delta)}{d \alpha} = -\frac{\left(12\alpha\delta^2+4\alpha^3\right)\cdot a_4+2\alpha a_2}{b} + \left(4\delta^3+12\alpha^2 \delta \right)\cdot a_4+2\delta \cdot a_2$$
Equate to zero and express $\alpha$:
$$\alpha^3 \cdot (2 \cdot a_4) - \alpha^2 \cdot (6 \cdot \delta \cdot a_4 \cdot b) + \alpha \cdot (6 \cdot \delta^2 \cdot a_4 + a_2) = 2 \cdot \delta^3 \cdot a_4 \cdot b + \delta \cdot a_2 \cdot b$$
Next, I can't find an analytic relation to calculate $\alpha$.
substituting $a_0=1; a_2 = -3/2; a_4 = 7/8; b = 2; \delta = 0.353$ I have to get the three roots of the equation.
Maybe you can set a condition that the search for the maximum is carried out within the required limits?
Hint.
With
$$f(\alpha, b,\delta) = \left(a_0 + a_2\cdot(\alpha-\delta)^2 + a_4\cdot(\alpha-\delta)^4\right)\cdot\left(1-\frac{b-1}{2b}\right)\cdot (-1) + \left(a_0 + a_2\cdot(\alpha+\delta)^2 + a_4\cdot(\alpha+\delta)^4\right)\cdot\frac{b-1}{2b}$$
introducing the new variables $u = \alpha + \delta, v = \alpha - \delta, w = \frac{b-1}{2b}$ we have
$$ g(u,v,w) = (w-1) \left(a_0+a_2 u^2+a_4 u^4\right)+w \left(a_0+a_2 v^2+a_4 v^4\right) $$
and after substitutions
$$ g_0(u,v) = \frac{1}{4} \left(\frac{7 v^4}{8}-\frac{3 v^2}{2}+1\right)-\frac{3}{4} \left(\frac{7 u^4}{8}-\frac{3 u^2}{2}+1\right) $$
The stationary points are determined by solving
$$ \nabla g_0(u,v) = 0 = \cases{\frac{3}{8} u \left(6-7 u^2\right)\\ \frac{1}{8} v \left(7 v^2-6\right)} $$
with solutions
$$ \left[ \begin{array}{ccc} g_0 & u & v\\ -\frac{37}{56} & 0 & -\sqrt{\frac{6}{7}} \\ -\frac{37}{56} & 0 & \sqrt{\frac{6}{7}} \\ -\frac{1}{2} & 0 & 0 \\ -\frac{5}{28} & -\sqrt{\frac{6}{7}} & -\sqrt{\frac{6}{7}} \\ -\frac{5}{28} & -\sqrt{\frac{6}{7}} & \sqrt{\frac{6}{7}} \\ -\frac{5}{28} & \sqrt{\frac{6}{7}} & -\sqrt{\frac{6}{7}} \\ -\frac{5}{28} & \sqrt{\frac{6}{7}} & \sqrt{\frac{6}{7}} \\ -\frac{1}{56} & -\sqrt{\frac{6}{7}} & 0 \\ -\frac{1}{56} & \sqrt{\frac{6}{7}} & 0 \\ \end{array} \right] $$