Question

expand 2x³+7x²+x-1 in power of (x-2) by taylor's theorom​

Answer 1

Answer:

Please find the attached file.

Step-by-step explanation:

Answer 2

Step-by-step explanation:

Given: $f(x)=2x^{3} +7x^{2} +x-1$

(Second half)

Considering $f(2),f'(2),f''(2)$ , and $f'''(2)$;

$=> f(2)=2(2)^{3} +7(2)^{2}+2-1\\=> f'(x)=6x^{2} +14x+1\\=>f'(2)=6(2)^{2} +14(2)+1=53\\=>f''(x)=12x +14=> f''(2)=12(2) +14=38\\=>f'''(x)=12=> f'''(2)=12$

Substituting $f(2),f'(2),f''(2)$ , and $f'''(2)$ in the (1), we get $f(x)=45+53(x-2)+\frac{38(x-2)^{2} }{2!} +\frac{12(x-2)^{3} }{3!}$  

Hence, Tayor's series expansion is

 $f(x)=45+53(x-2)+\frac{38(x-2)^{2} }{2!} +\frac{12(x-2)^{3} }{3!}$

Note: There is no swear words but still it didn't get upload so please refer to the picture for first half of the solution