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EastWind [94]
4 years ago
12

Find f(−1) given that f(x)=x3+2x2+3x+4. A. 10 B. -2 C. 2 D. -10

Mathematics
1 answer:
Finger [1]4 years ago
8 0

Answer:

The answer is option C

Step-by-step explanation:

f(x) = x³ + 2x² + 3x + 4

To find f(-1) substitute the value of x that's

- 1 into f(x)

That's

<h3>f( - 1) =  ({ - 1})^{3}  + 2 ({ - 1})^{2}  + 3( - 1) + 4 \\  =  - 1 + 2(1)  - 3 + 4 \\  =  - 1 + 2 - 3 + 4 \\  = 1 - 3 + 4</h3>

We have the final answer as

<h3>f( - 1) = 2</h3>

Hope this helps you

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\displaystyle\\(2x+2)^6=\frac{6!}{(6-0)!*0!} (2x)^62^0+\frac{6!}{(6-1)!*1!} (2x)^{6-1}2^1+\frac{6!}{(6-2)!*2!}(2x)^{6-2}2^2+\\\\ +\frac{6!}{(6-3)!*3!} (2a)^{6-3}2^3+\frac{6!}{(6-4)*4!} (2x)^{6-4}b^4+\frac{6!}{(6-5)!*5!}(2x)^{6-5} b^5+\frac{6!}{(6-6)!*6!}(2x)^{6-6}b^6. \\\\

(2x+2)^6=\frac{6!}{6!*1} 2^6*x^6*1+\frac{5!*6}{5!*1}2^5*x^5*2+\\\\+\frac{4!*5*6}{4!*1*2}2^4*x^4*2^2+  \frac{3!*4*5*6}{3!*1*2*3} 2^3*x^3*2^3+\frac{4!*5*6}{2!*4!}2^2*x^2*2^4+\\\\+\frac{5!*6}{1!*5!} 2^1*x^1*2^5+\frac{6!}{0!*6!} x^02^6\\\\(2x+2)^6=64x^6+384x^5+960x^4+1280x^3+960x^2+384x+64.

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