Question

Please help
Find the set of values taken by 1/ (9x^2+12x+7) for real values of x

Answer 1

`\textsf{Range:}\quad \left(0,(1)/(3)\right]`

Explanation:

To find the set of values taken by a function, we need to determine the range of the function.

Given rational function:

`f(x)=(1)/(9x^2+12x+7)`

Vertical asymptotes occur where the denominator of the rational function is equal to zero. Therefore, we can solve the quadratic equation 9x² + 12x + 7 = 0 to find any vertical asymptotes.

In this case, the quadratic equation has complex roots, and therefore, there are no real values of x for which the denominator is zero. So, there are no vertical asymptotes.

Horizontal asymptotes can be determined by the relationship between the degrees of the numerator and denominator. In this case, the degree of the numerator is less than the degree of the denominator, which indicates that there is a horizontal asymptote at y = 0.

Upon examining the end behavior of the function, as x approaches positive or negative infinity, the function approaches the horizontal asymptote y = 0 from the positive y-axis.

Critical points of a function are values in its domain where the derivative is either zero or undefined, and they correspond to potential locations of local extrema or points of inflection. Therefore, to find the local extrema, differentiate the function using the quotient rule.

`f(x)=(1)/(9x^2+12x+7)`

`\textsf{Let}\;\;u=1 \implies \frac{\text{d}u}{\text{d}x}=0`

`\textsf{Let}\;\;v=9x^2+12x+7 \implies \frac{\text{d}v}{\text{d}x}=18x+12`

`f'(x)=\frac{v\frac{\text{d}u}{\text{d}x}-u\frac{\text{d}v}{\text{d}x}}{v^2}`

`f'(x)=((9x^2+12x+7)(0)-(1)(18x+12))/((9x^2+12x+7) ^2)`

`f'(x)=-(18x+12)/((9x^2+12x+7) ^2)`

Set f'(x) equal to zero and solve for x:

`-(18x+12)/((9x^2+12x+7) ^2)=0`

`-(18x+12)=0`

`18x+12=0`

`x=-(12)/(18)`

`x=-(2)/(3)`

Substitute x = -2/3 into the function to find the corresponding value of y:

`f\left(-(2)/(3)\right)=(1)/(9\left(-(2)/(3)\right)^2+12\left(-(2)/(3)\right)+7)=(1)/(3)`

Therefore, there is a local extrema at (-2/3, 1/3). As we have already established that the range of the function is positive only, and the function approaches y = 0 as x approaches positive and negative infinity, point (-2/3, 1/3) must be the local maximum.

In conclusion, considering the behavior of the function, we can state that the range of the function is all real numbers greater than zero and equal to or less than 1/3:

`\large\boxed{\boxed{\textsf{Range:}\quad \left(0,(1)/(3)\right]}}`