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

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(B) The given function is a geometric series with first term $a = 1$,common ratio $r = 2x^2$,and $n = 11$ terms.$f(x) = \sum_{k=0}^{10} (2x^2)^k = \fr

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Exactly one minimum

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(B) The given function is a geometric series with first term $a = 1$,common ratio $r = 2x^2$,and $n = 11$ terms.$f(x) = \sum_{k=0}^{10} (2x^2)^k = \frac{(2x^2)^{11} - 1}{2x^2 - 1}$.Alternatively,we can differentiate the polynomial directly:$f(x) = 1 + 2x^2 + 4x^4 + 8x^6 + \dots + 1024x^{20}$.$f'(x) = 4x + 16x^3 + 48x^5 + \dots + 20480x^{19}$.$f'(x) = 4x(1 + 4x^2 + 12x^4 + \dots + 5120x^{18})$.Setting $f'(x) = 0$,we see that $x = 0$ is the only real critical point because the term in the bracket is always positive for all real $x$.Calculating the second derivative at $x = 0$:$f''(x) = 4 + 48x^2 + 240x^4 + \dots$$f''(0) = 4 > 0$.Since $f'(0) = 0$ and $f''(0) > 0$,the function has a local minimum at $x = 0$. Since it is a polynomial of even degree with a positive leading coefficient,this is the global minimum. Thus,$f(x)$ has exactly one minimum.

Let $f(x) = 1 + 2x^2 + 2^2x^4 + \dots + 2^{10}x^{20}$. Then $f(x)$ has:

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