Find all possible values ( range) of the following quadratic expressions when `x in R` and when `x in [-3,2]`
(a) `4x^2+28x+41`
(b) `1+6x -x^2`

Explanation

To solve the given problem, we will find the range of the quadratic expressions \(4x^2 + 28x + 41\) and \(1 + 6x - x^2\) for \(x \in \mathbb{R}\) and for \(x \in [-3, 2]\).

### Part (a): Finding the range of \(4x^2 + 28x + 41\)

1. **Rewrite the expression**:
We can complete the square for the quadratic expression:
\[ 4x^2 + 28x + 41 = 4(x^2 + 7x) + 41 \]
To complete the square, we take half of the coefficient of \(x\) (which is 7), square it, and adjust:
\[ = 4\left(x^2 + 7x + \left(\frac{7}{2}\right)^2 - \left(\frac{7}{2}\right)^2\right) + 41 \]
\[ = 4\left(\left(x + \frac{7}{2}\right)^2 - \frac{49}{4}\right) + 41 \]
\[ = 4\left(x + \frac{7}{2}\right)^2 - 49 + 41 \]
\[ = 4\left(x + \frac{7}{2}\right)^2 - 8 \]

2. **Determine the range for \(x \in \mathbb{R}\)**:
The term \(4\left(x + \frac{7}{2}\right)^2\) is always non-negative (since it is a square), and its minimum value is 0 when \(x = -\frac{7}{2}\).
Therefore, the minimum value of the entire expression is:
\[ -8 \]
Thus, the range for \(x \in \mathbb{R}\) is:
\[ \text{Range} = [-8, \infty) \]

3. **Determine the range for \(x \in [-3, 2]\)**:
We evaluate the function at the endpoints and check for critical points within the interval.
- At \(x = -3\):
\[ f(-3) = 4(-3)^2 + 28(-3) + 41 = 36 - 84 + 41 = -7 \]
- At \(x = 2\):
\[ f(2) = 4(2)^2 + 28(2) + 41 = 16 + 56 + 41 = 113 \]
- The vertex occurs at \(x = -\frac{7}{2} = -3.5\), which is outside the interval \([-3, 2]\).

Therefore, the maximum value in the interval is \(113\) and the minimum is \(-7\).
Thus, the range for \(x \in [-3, 2]\) is:
\[ \text{Range} = [-7, 113] \]

### Part (b): Finding the range of \(1 + 6x - x^2\)

1. **Rewrite the expression**:
We can rewrite the quadratic expression:
\[ 1 + 6x - x^2 = -x^2 + 6x + 1 \]
To find the vertex, we can complete the square:
\[ = -\left(x^2 - 6x\right) + 1 \]
Completing the square:
\[ = -\left((x - 3)^2 - 9\right) + 1 \]
\[ = - (x - 3)^2 + 9 + 1 \]
\[ = - (x - 3)^2 + 10 \]

2. **Determine the range for \(x \in \mathbb{R}\)**:
The maximum value occurs at the vertex \(x = 3\):
\[ f(3) = 10 \]
Since the parabola opens downwards, the range is:
\[ \text{Range} = (-\infty, 10] \]

3. **Determine the range for \(x \in [-3, 2]\)**:
We evaluate the function at the endpoints and check for critical points within the interval.
- At \(x = -3\):
\[ f(-3) = 1 + 6(-3) - (-3)^2 = 1 - 18 - 9 = -26 \]
- At \(x = 2\):
\[ f(2) = 1 + 6(2) - (2)^2 = 1 + 12 - 4 = 9 \]
- The vertex \(x = 3\) is outside the interval \([-3, 2]\).

Therefore, the maximum value in the interval is \(9\) and the minimum is \(-26\).
Thus, the range for \(x \in [-3, 2]\) is:
\[ \text{Range} = [-26, 9] \]

### Summary of Ranges
- For \(4x^2 + 28x + 41\):
- \(x \in \mathbb{R}\): \([-8, \infty)\)
- \(x \in [-3, 2]\): \([-7, 113]\)

- For \(1 + 6x - x^2\):
- \(x \in \mathbb{R}\): \((-\infty, 10]\)
- \(x \in [-3, 2]\): \([-26, 9]\)