Question

range of f(x) =3/2-x2.is

Answer 1

Hey there!

For a function range or a range function the set of obtained values are of that to the dependent variable for which that function is defined.

Firstly find the vertex of this function that is,

Vertex of $\bf{\dfrac{3}{2} - x^2}$ :

Write this in the standard form of $\bf{y = ax^2 + bx + c}$ :

$\bf{y = - x^2 + \dfrac{3}{2}}$

Now, for a given parabola of $\bf{ax^2 + bx + c}$ the vertex of which "x" equals to the fractional variable of $\bf{\dfrac{- b}{2a}}$; Here, a = - 1,  and,  b = 0.

$\bf{\therefore \quad x = \dfrac{- 0}{2(- 1)}}$

$\bf{\therefore \quad x = 0}$

Now, Just plug or put in the value of variable "x" to find the value of variable "y":

$\bf{\therefore \quad y = - 0^2 + \dfrac{3}{2}}$

$\boxed{\bf{\underline{\therefore \quad y = \dfrac{3}{2}}}}$

Therefore, the parabola of this vertex becomes:

$\boxed{\bf{\underline{(0, \: \: \dfrac{3}{2})}}}$

Now, the conditional variable set values of "a" is as follows;

If a is lesser than the value of "0" then the vertex value is said to be a maximum value.

If a is greater than the value of "0" then the vertex value is said to be a minimum value.

Here, a = -1.

Therefore, the vertex value is a maximum value.

$\boxed{\bf{\underline{Maximum \: \: \: \: (0, \: \: \dfrac{3}{2})}}}$

For a parabola in the equation of $\bf{ax^2 + bx + c}$ with a specified vertex of $\bf{(x_v, \: y_v)}$.

Now, If a is lesser than the value of "0" then variable "a" is in the range of $\bf{f(x) \leq y_v}$.

And, if a is greater than the value of "0" then variable "a" is in the range of $\bf{f(x) \geq y_v}$.

Here,  a = - 1.  Vertex of this that is,

$\boxed{\bf{\underline{(x_v, \: y_v) = (0, \: \: \dfrac{3}{2})}}}$

$\boxed{\bf{\underline{\therefore \quad f(x) \leq \dfrac{3}{2}}}}$

Interval Notation for this functional range vertex is,

$\boxed{\bf{\underline{\therefore \quad Interval \: \: Notation \: = (- \infty, \: \: \dfrac{3}{2}]}}}$

Which is the required solution for this type of query.

[For Graph Check the attachment]

Hope this helps you and solves your doubts for function range!!!!