解不等式 $2y - 1 < \frac{11y}{4} < \frac{1}{4}$ 。答案：... 已知 -25 ≤ x ≤ 50 且 -15 ≤ y ≤ -5，x + y^2 的最小可能值是多少？答案：...

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完整解法

1. Split the compound inequality into two inequalities.

$2y-1<\frac{11y}{4}\quad \text{and}\quad \frac{11y}{4}<\frac14$

2. Solve the left inequality.

$2y-1<\frac{11y}{4}$
Multiply by $4$ :
$8y-4<11y$
$-4<3y$
$y>-\frac43$

3. Solve the right inequality and combine both conditions.

$\frac{11y}{4}<\frac14$
Multiply by $4$ :
$11y<1$
$y<\frac{1}{11}$

Combining with $y>-\frac43$ gives
$-\frac43<y<\frac{1}{11}.$

答案

$0<y<\dfrac{4}{7}$

1. Find the minimum possible values contributing to $x+y^2$.

Since $-25\leq x\leq 50$ , the smallest possible value of $x$ is $-25$ .

Since $-15\leq y\leq -5$ , the smallest possible value of $y^2$ occurs at $y=-5$ :
$y^2=(-5)^2=25.$

2. Calculate the smallest possible value of the expression.

Smallest possible value:
$x+y^2=-25+25=0.$

答案

$0$

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