How do you find the domain and range of #(x-3) (3x-2)#?

Answer 1

Both the domain and the range are #-oo# to #+oo#.

The function is #= 3x^2 - 11x +6#
The domain is defined as any value of #x# that if you plug in the eq, shall give you a real value of the function.
So if you plug any value of #x# above, you will get a real value.

The range is defined as the set of values obtained by plugging in the values of the domain of the function.

Thus, domain and range are both are all real numbers.

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Answer 2

To find the domain and range of the function ( f(x) = (x - 3)(3x - 2) ), we consider the restrictions on the values of ( x ) that make the function undefined and the possible output values of the function.

Domain: The domain of ( f(x) ) consists of all real numbers for which the function is defined. In this case, there are no restrictions on the values of ( x ), so the domain is all real numbers, or ((-∞, +∞)).

Range: To find the range, we need to determine the possible output values of the function. Since ( f(x) ) is a quadratic function, it will have a range that depends on the behavior of the quadratic expression ( (x - 3)(3x - 2) ).

To analyze the behavior of the quadratic expression, we can look at its leading coefficient and its concavity. The leading coefficient is positive (( 3 )), indicating that the quadratic expression opens upwards, and its concavity does not change. Therefore, the range of ( f(x) ) is all real numbers greater than or equal to the minimum value of the quadratic expression.

To find the minimum value of the quadratic expression, we can use its vertex formula. The vertex of a quadratic function ( ax^2 + bx + c ) is given by ( \left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right) ).

In this case, the quadratic expression ( (x - 3)(3x - 2) ) can be expanded and then its vertex can be found. After finding the vertex, we can determine the minimum value and conclude that the range of ( f(x) ) is all real numbers greater than or equal to that minimum value.

I will calculate it for you.

The expression ( (x - 3)(3x - 2) ) can be expanded to ( 3x^2 - 11x + 6 ). The vertex of this quadratic function is at ( x = \frac{-b}{2a} = \frac{11}{6} ).

Substituting ( x = \frac{11}{6} ) into the function, we find: [ f\left(\frac{11}{6}\right) = 3\left(\frac{11}{6}\right)^2 - 11\left(\frac{11}{6}\right) + 6 = \frac{1}{12} ]

Therefore, the minimum value of the function is ( \frac{1}{12} ), and the range of ( f(x) ) is all real numbers greater than or equal to ( \frac{1}{12} ), or ([ \frac{1}{12}, +∞) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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