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katrin2010 [14]
3 years ago
8

What equals the product of (x-3)(2x+1)

Mathematics
1 answer:
melisa1 [442]3 years ago
3 0
(x-3)(2x+1) 

 <span>=<span><span><span><span><span>(x)</span><span>(<span>2x</span>)</span></span>+<span><span>(x)</span><span>(1)</span></span></span>+<span><span>(<span>−3</span>)</span><span>(<span>2x</span>)</span></span></span>+<span><span>(<span>−3</span>)</span><span>(1)</span></span></span></span><span>=<span><span><span><span>2<span>x^2</span></span>+x</span>−<span>6x</span></span>−3</span></span><span>=<span><span><span>2<span>x^2</span></span>−<span>5x</span></span>−<span>3</span></span></span>
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Answer:

Part A.

Let f(x) = 0;

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f(a) + hf'±(a) + \frac{h^{2} }{2!}f''(a) = 0

h = \frac{-f'(a) }{f''(a)} ± \frac{\sqrt[]{(f'(a))^{2}-2f(a)f''(a) } }{f''(a)}

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Part B.

It is evident that Newton's method fails in two cases, as:

1.  if f''(x) = 0

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Part C.

In case  x_{i+1} is close to x_{i}, the choice that shouldbe made instead of ± in part A is:

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Part D.

As given x_{i+1} = x_{i} = h

or                 h = x_{i+1} - x_{i}

We get,

f(a) + hf'(a) +(h²/2)f''(a) = 0

or h² = -hf(a)/f'(a)

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So,                f(a) + hf'(a) - (f''(a)/2)(hf(a)/f'(a)) = 0

It becomes   h = -f(a)/f'(a) + (h/2)[f''(a)f(a)/(f(a))²]

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3 years ago
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Answer:

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