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2 Answers

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  1. Minima at -0.5

First derivative to 0 gives x = -0.5

Second derivative gives 2, as its positive, -0.5 is minima.

Cannot be saddle point because it's a quadratic function and graph will have only 1 minima

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To determine whether the critical point \(x = -0.5\) corresponds to a minimum, maximum, or saddle point for the function \(f(x) = 1 + x + x^2\), we can use the second derivative test.

The second derivative of \(f(x)\) is the derivative of the first derivative:

\[f'(x) = 1 + 2x\]

Now, evaluate \(f''(-0.5)\):

\[f''(-0.5) = 2\]

The second derivative is positive, indicating that the function has a local minimum at \(x = -0.5\) (since the concavity changes from negative to positive).

Therefore, the correct answer is:
- Minima at \(x = -0.5\)

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