For what values of x is #f(x)=-x^3+x^2-x+5# concave or convex?

Answer 1

Second derivative is positive and hence it is convex or minimum.

#f’(x) = -3x^2 + 2x -1#
#f”(x) = -6x + 2 = 0# # 6x = 2 or x = 1/3#

Therefore it is convex or minimum.

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Answer 2
To determine where the function \( f(x) = -x^3 + x^2 - x + 5 \) is concave or convex, you need to find the second derivative of the function and then analyze its sign. 1. Find the first derivative of \( f(x) \): \[ f'(x) = -3x^2 + 2x - 1 \] 2. Find the second derivative of \( f(x) \): \[ f''(x) = -6x + 2 \] 3. Set \( f''(x) \) equal to zero and solve for \( x \) to find the points of inflection: \[ -6x + 2 = 0 \] \[ x = \frac{1}{3} \] 4. Analyze the sign of \( f''(x) \) in the intervals separated by the point of inflection, \( x = \frac{1}{3} \): - For \( x < \frac{1}{3} \), \( f''(x) \) is positive, so the function is concave up. - For \( x > \frac{1}{3} \), \( f''(x) \) is negative, so the function is concave down. Therefore, the function \( f(x) = -x^3 + x^2 - x + 5 \) is concave up for \( x < \frac{1}{3} \) and concave down for \( x > \frac{1}{3} \).
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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