Final Answer:
The absolute minimum value of g(x) = -²x² - ²x - ¹ over the interval
is c) -¹⁷.
Step-by-step explanation:
To find the absolute minimum of the quadratic function g(x) = -²x² - ²x - ¹, we first locate its critical points by finding where the derivative is zero. Taking the derivative and setting it equal to zero gives us g'(x) = -4x - 2 = 0. Solving for x, we find x = -½, which is the only critical point.
To determine whether this critical point corresponds to a minimum, maximum, or neither, we examine the sign of the second derivative. Calculating g''(x), we get g''(x) = -4. Since the second derivative is negative, the critical point x = -½ corresponds to a local maximum.
To find the absolute minimum over the given interval
]
], we evaluate g(x) at the endpoints and the critical point. g(-3) = -14, g(-½) = -17/4, and g(4) = -33. The smallest of these values is -17/4, which corresponds to option c) -¹⁷. Therefore, -¹⁷ is the absolute minimum of g(x) over the interval
![[-3,4][-3 , 4 ].](https://img.qammunity.org/2024/formulas/mathematics/high-school/3q5t9gi4gyc2afkz219km6e6et8jykic46.png)