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3 years ago

6

If f(x) =x/2 -2 and g(x) = 2x2 + x - 3, find (8 +g)(x),

2 answers:

anastassius [24]3 years ago

4 0

Answer: D) $2x^{2} +\frac{3x}{2} -5$

Step-by-step explanation:

$f(x)=\frac{x}{2} -2 \\g(x)= 2x^{2} +x-3\\\\\(f+g)(x)\\\$

$\frac{x}{2} -2 +2x^{2} +x-3\\2x^{2} +\frac{x}{2} +x-5\\\\(f+g)(x)= 2x^{2} +\frac{3x}{2} -5$

vesna_86 [32]3 years ago

3 0

I would go with D. 2x^2+3/2x-5

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pickupchik [31]

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Step-by-step explanation:

The Newton's Method is a multi-step numerical method for continuous diffentiable function of the form $f(x) = 0$ based on the following formula:

$x_{i+1} = x_{i} -\frac{f(x_{i})}{f'(x_{i})}$

Where:

$x_{i}$ - i-th Approximation, dimensionless.

$x_{i+1}$ - (i+1)-th Approximation, dimensionless.

$f(x_{i})$ - Function evaluated at i-th Approximation, dimensionless.

$f'(x_{i})$ - First derivative evaluated at (i+1)-th Approximation, dimensionless.

Let be $f(x) = x^{2}-8$ and $f'(x) = 2\cdot x$ , the resultant expression is:

$x_{i+1} = x_{i} -\frac{x_{i}^{2}-8}{2\cdot x_{i}}$

First iteration: ( $x_{1} = 2$ )

$x_{2} = 2-\frac{2^{2}-8}{2\cdot (2)}$

$x_{2} = 2 + \frac{4}{4}$

$x_{2} = 3$

Second iteration: ( $x_{2} = 3$ )

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3 years ago