Find the smallest integer value of $c$ such that the function $f(x)=\frac{2x^2+x+5}{x^2+4x+c+12x+x^2}$ has a domain of all real numbers.
The function $f(x)$ is defined for all real numbers except for those that make the denominator equal to zero. Hence we need $x^2+4x+c+12x+x^2=0$. Simplifying gives $2x^2+16x+c=0$. This polynomial is a quadratic, which means the graph of $y=2x_{^2}+16x+c$ is a parabola.
We are looking for the smallest integer value of $c$ such that this parabola has no real roots. There are a few ways to see what values of $c$ this is true for. For example, we can complete the square: $2x^2 + 16x + c = 2(x^2+8x) + c = 2((x+4)^2-16)+c = 2(x+4)^2 - 32 + c$. Since the square of a real number is nonnegative, the smallest possible value of $2(x+4)^2$ is 0, and adding 0 to $-32$ will not change its value. Therefore, the smallest integer value of $c$ such that this parabola has no real roots is $c=\boxed{-32}$.
To find the smallest integer value of $c$ such that the function $f(x)$ has a domain of all real numbers, we need to consider the denominator of the function.
The denominator is given by $x^2+4x+c+12x+x^2$. To find the domain, we need to make sure that the denominator is not equal to zero.
Setting the denominator equal to zero, we have $(x^2 + x^2) + (4x + 12x) + c = 0$.
Combining like terms, we have $2x^2 + 16x + c = 0$.
For this quadratic equation to have real solutions for $x$, the discriminant must be non-negative. The discriminant is given by $b^2 - 4ac$, where $a$, $b$, and $c$ are the coefficients of the quadratic equation.
In this case, $a = 2$, $b = 16$, and $c$ is the value we want to find. So we have $16^2 - 4(2)(c) \geq 0$.
Simplifying, we have $256 - 8c \geq 0$.
Adding $8c$ and then dividing both sides by $8$, we have $256 \geq 8c$.
Dividing both sides by $8$, we get $32 \geq c$.
Therefore, the smallest integer value of $c$ that satisfies the condition is $c = 32$.
To find the smallest integer value of $c$ such that the function has a domain of all real numbers, we need to determine the values of $x$ for which the denominator of the function is not equal to zero.
The denominator of the function is $x^2+4x+c+12x+x^2 = 2x^2+16x+c$. For the function to have a domain of all real numbers, the quadratic expression $2x^2+16x+c$ should never be equal to zero.
For a quadratic equation $ax^2+bx+c=0$, the discriminant is given by $b^2-4ac$. If the discriminant is greater than or equal to zero, the quadratic equation has real roots, which means that the quadratic expression does not equal zero for all real values of $x$.
In this case, the discriminant of $2x^2+16x+c$ is $16^2-4(2)(c) = 256-8c$. For the function to have a domain of all real numbers, the discriminant should be greater than or equal to zero, so we have $256-8c \geq 0$.
Solving this inequality, we get $c \leq 32$. Therefore, the smallest integer value of $c$ such that the function has a domain of all real numbers is $c = \boxed{32}$.