Let f(x)=\(x-\frac{x^3}{3^2.2!}+\frac{x^5}{5^2.4!}-\frac{x^7}{7^2.6!}+…\infty\). Find a point nearest to c such that f'(c) = 1

Let f(x)=\(x-\frac{x^3}{3^2.2!}+\frac{x^5}{5^2.4!}-\frac{x^7}{7^2.6!}

1.	Let f(x)=\(x-\frac{x^3}{3^2.2!}+\frac{x^5}{5^2.4!}-\frac{x^7}{7^2.6!}+…\infty\). Find a point nearest to c such that f'(c) = 1
A.	1
B.	0
C.	= 1a) 1b) 0c) 2.3445 * 10-9
D.	458328.33 * 10-3view answer
Answer» E.

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Let f(x)=\(x-\frac{x^3}{3^2.2!}+\frac{x^5}{5^2.4!}-\frac{x^7}{7^2.6!}+…\infty\). Find a point nearest to c such that f'(c) = 1

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