If f(x) = 1 + 2x^2 + 2^2 x^4 + dots + 2^{10} | vedclass

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(B) Given $f(x) = 1 + 2x^2 + 2^2 x^4 + \dots + 2^{10} x^{20}$.This is a geometric progression with $a = 1$,common ratio $r = 2x^2$,and $n = 11$ terms.

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Exactly one minimum.

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(B) Given $f(x) = 1 + 2x^2 + 2^2 x^4 + \dots + 2^{10} x^{20}$.This is a geometric progression with $a = 1$,common ratio $r = 2x^2$,and $n = 11$ terms.$f(x) = \frac{1((2x^2)^{11} - 1)}{2x^2 - 1} = \frac{(2x^2)^{11} - 1}{2x^2 - 1}$.Alternatively,differentiating $f(x)$ directly:$f'(x) = 0 + 2(2x) + 2^2(4x^3) + \dots + 2^{10}(20x^{19})$.$f'(x) = 4x + 16x^3 + \dots + 2^{10}(20x^{19}) = 4x(1 + 2(2x^2) + 3(2x^2)^2 + \dots + 10(2x^2)^9)$.Setting $f'(x) = 0$,we get $x = 0$ as the only real critical point because the term in the bracket is a sum of positive terms for all $x 
eq 0$ (specifically,it is the derivative of a geometric series which is strictly increasing for $x^2 > 0$).For $x  0$,$f'(x) > 0$.By the first derivative test,$f(x)$ has a local minimum at $x = 0$.Since $x=0$ is the only critical point,$f(x)$ has exactly one minimum.

If $f(x) = 1 + 2x^2 + 2^2 x^4 + \dots + 2^{10} x^{20}$,then $f(x)$ has:

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