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Consider the function \( f(x) = 2x^2 + 71 \).

(a) \( f \) is concave up for \( x \in \)

(b) \( f \) is concave down for \( x \in \)

(c) The inflection points of \( f \) occur at \( x = \)

Answer :

f is concave up for x∈ (−∞, +∞).

  • f is concave down for x∈ (−∞, +∞).

  • The inflection points of f occur at x = None.

A function is concave up when its second derivative is positive. Taking the second derivative of f(x) = 2x^2 + 71, we get f''(x) = 4. Since f''(x) is a constant value of 4, which is positive, it indicates that the function is concave up for all real values of x.

A function is concave down when its second derivative is negative. In the case of f(x) = 2x^2 + 71, the second derivative f''(x) is equal to 4, which is positive. Since f''(x) is positive and not negative, the function is not concave down for any values of x.

An inflection point occurs when the concavity of a function changes. In the case of f(x) = 2x^2 + 71, the concavity remains the same (concave up) for all values of x because the second derivative, f''(x), is always positive. Therefore, there are no inflection points in this function.

It's important to note that in this particular case, the function f(x) = 2x^2 + 71 is always concave up and does not have any inflection points.

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