The absolute minimum value of g(x) on the closed interval [−2,2] is 0, which occurs at x= −2.
To find the absolute minimum value of the function
g(x)=∣x∣+2∣1−x∣ on the closed interval [−2,2], we need to consider three different cases for the absolute value expressions.
When
x is non-negative (x≥0), the function simplifies to g(x)=x+2(1−x)=−x+2.
In this case, the minimum occurs at x=0, and g(0)=2.
When
x is negative (x<0), the function becomes g(x)=−x+2−2(1−x)=x+2.
Here, the minimum occurs at the endpoint x=−2, and g(−2)=0.
When x is positive (x>0), the function is g(x)=x+2(1−x)=−x+2.
Similar to case 1, the minimum occurs at x=0, and g(0)=2.
Now, we compare the minimum values obtained in each case: g(−2)=0, g(0)=2, and g(0)=2.
The absolute minimum value of g(x) on the closed interval [−2,2] is 0, which occurs at x=−2.