Question

Image set p=x^2+1 (-2<=x<=2)?

Answer 1

The Right Option is `C [1,5]`.

Explanation:

`-2 le x le 2 rArr 0 le x^2 le 4`.

Adding `1, 1 le x^2+1 le 5`.

`:. AA x in [-2,2], p(x)=x^2+1 in [1,5]`, which is the desired

Image Set.

Hence, the right option is `C [1,5]`.

Answer 2

`C [1,5]`

Explanation:

.

The image of a function is all the possible values a function can have in a specified domain.

Your specified domain is `-2 < x < 2`. We have to check and see what values `p(x)` can have between `-2` and `2`.

`p(x)=x^2+1`

This is a quadratic function, i.e. it is a parabola. Since the coefficient of `x^2` is positive, the parabola opens upward. This means that the vertex of the parabola is where the function has its minimum value.

If the function of your parabola is given in the form of:

`y=ax^2+bx+c` then the `x`-coordinate of the vertex can be calculated as follows:

`x_(vertex)=(-b)/(2a)`

In this problem, `a=1` and `b=0` because there is no `x` term.

`x_(vertex)=(-b)/(2a)=-0/(2(1))=0`

Now, we can plug this value into the function to find the `y`-coordinate of the vertex:

`y_(vertex)=(0)^2+1=1`

This means that at `x=0` the value of your function is `1` and this is the minimum value of your function between `-2` and `2` as we can see from the graph below:

graph{x^2+1 [-10, 10, -5, 5]}

But the maximum value of your function would be `oo` if your domain were not restricted. Since your domain is given as:

`-2 < x < 2`

let's find the value of your function at `-2` and `2`:

`p(-2)=(-2)^2+1=4+1=5`

`p(2)=(2)^2+1=4+1=5`

So, the maximum value of the function in this domain is `5` and the minimum value is `1` which is why option `C` is correct.