Let f(x)=1+frac{x}{1 !}+frac{x^2}{2 !}+frac{x | vedclass

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(A) We have $f(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}$.Note that $f'(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} = f(x) - \fra

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(A) We have $f(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}$.Note that $f'(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} = f(x) - \frac{x^4}{4!}$.Also,$f''(x) = 1 + x + \frac{x^2}{2} = \frac{1}{2}(x^2 + 2x + 1) + \frac{1}{2} = \frac{1}{2}(x+1)^2 + \frac{1}{2} > 0$ for all $x \in \mathbb{R}$.Since $f''(x) > 0$,$f'(x)$ is a strictly increasing function.As $x 	o -\infty$,$f'(x) 	o -\infty$ and as $x 	o \infty$,$f'(x) 	o \infty$. Thus,$f'(x) = 0$ has exactly one real root,say $x_0$.Since $f'(x)$ is increasing,$f(x)$ has a global minimum at $x = x_0$.We observe $f'(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$.$f'(-2) = 1 - 2 + 2 - \frac{8}{6} = 1 - 1.33 = -0.33 $f'(-1) = 1 - 1 + \frac{1}{2} - \frac{1}{6} = \frac{1}{3} > 0$.So,$x_0 \in (-2, -1)$.At the minimum $x_0$,$f(x_0) = 1 + x_0 + \frac{x_0^2}{2} + \frac{x_0^3}{6} + \frac{x_0^4}{24}$.Since $f'(x_0) = 0$,we have $1 + x_0 + \frac{x_0^2}{2} + \frac{x_0^3}{6} = 0$,which implies $1 + x_0 + \frac{x_0^2}{2} + \frac{x_0^3}{6} = -\frac{x_0^4}{24}$.Substituting this into $f(x_0)$,we get $f(x_0) = -\frac{x_0^4}{24} + \frac{x_0^4}{24} + \dots$ wait,$f(x_0) = (1 + x_0 + \frac{x_0^2}{2} + \frac{x_0^3}{6}) + \frac{x_0^4}{24} = 0 + \frac{x_0^4}{24} = \frac{x_0^4}{24}$.Since $x_0 
eq 0$,$f(x_0) = \frac{x_0^4}{24} > 0$.Since the global minimum of $f(x)$ is positive,$f(x) = 0$ has no real roots.

Let $f(x)=1+\frac{x}{1 !}+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\frac{x^4}{4 !}$. The number of real roots of $f(x)=0$ is

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