f(x)=x^3+6x^2+9x
find:
x intercepts
vertical asymptotes
horizontal/slant asymptotes
derivative
second derivative
increasing interval
decreasing interval
concave up interval
concave down interval
To find the x-intercepts of a function, you need to set f(x) equal to zero and solve for x. In this case, f(x) = x^3 + 6x^2 + 9x. Therefore, we set the equation equal to zero:
x^3 + 6x^2 + 9x = 0
Since there are no common factors to factor out, we can apply the zero product property and set each factor equal to zero:
x(x^2 + 6x + 9) = 0
To solve the equation x^2 + 6x + 9 = 0, we can either factor or use the quadratic formula. In this case, the quadratic factors into a perfect square:
(x + 3)^2 = 0
Taking the square root of both sides:
x + 3 = 0
Solving for x gives:
x = -3
Thus, the x-intercept of the function f(x) is -3.
To find the vertical asymptotes, we examine the behavior of the function as x approaches positive or negative infinity. In this case, the degree of the numerator is equal to the degree of the denominator (both are 3). Hence, there will not be any vertical asymptotes.
The function has no horizontal or slant asymptotes because the degrees of the numerator and denominator are equal.
To find the derivative of the function f(x), we can apply the power rule and the sum rule of differentiation. The power rule states that the derivative of x^n is n * x^(n-1). In this case, applying the power rule for each term:
f'(x) = 3x^2 + 12x + 9
To find the second derivative, we differentiate f'(x) using the power rule:
f''(x) = 6x + 12
To determine the increasing and decreasing intervals of the function, we need to examine the sign of the derivative. We set the derivative f'(x) equal to zero and solve for x:
3x^2 + 12x + 9 = 0
The quadratic equation does not factor easily, so we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
In this case, a = 3, b = 12, and c = 9.
x = (-12 ± √(12^2 - 4*3*9))/(2*3)
Simplifying:
x = (-12 ± √(144 - 108))/(6)
x = (-12 ± √(36))/(6)
x = (-12 ± 6)/(6)
x = -3 or -1
These are the critical points of the function. We can use test points or the first derivative test to determine the intervals of increasing and decreasing. Testing between -∞ and -3, we plug in a value like -4:
f'(-4) = 3(-4)^2 + 12(-4) + 9 = 12 > 0
Since f'(-4) is positive, the function is increasing in this interval.
Testing between -3 and -1, we plug in a value like -2:
f'(-2) = 3(-2)^2 + 12(-2) + 9 = -3 < 0
Since f'(-2) is negative, the function is decreasing in this interval.
To determine the concave up and concave down intervals, we need to examine the sign of the second derivative. The second derivative f''(x) is equal to 6x + 12. Setting f''(x) equal to zero and solving for x:
6x + 12 = 0
x = -2
Since f''(-2) = 0, we can use test points or the second derivative test to determine the intervals of concavity. Testing to the left of -2, we plug in a value like -3:
f''(-3) = 6(-3) + 12 = -6 < 0
Since f''(-3) is negative, the function is concave down in this interval.
Testing to the right of -2, we plug in a value like -1:
f''(-1) = 6(-1) + 12 = 6 > 0
Since f''(-1) is positive, the function is concave up in this interval.
In summary:
- The x-intercept of the function f(x) = x^3 + 6x^2 + 9x is -3.
- There are no vertical asymptotes.
- There are no horizontal or slant asymptotes.
- The first derivative of f(x) is f'(x) = 3x^2 + 12x + 9.
- The second derivative of f(x) is f''(x) = 6x + 12.
- The function is increasing on the interval (-∞, -3) and decreasing on the interval (-3, -1).
- The function is concave down on the interval (-∞, -2) and concave up on the interval (-2, +∞).