2 Answers | Add Yours
We have to show that 2x^3+12x^2+18x+24 = 0 has one real root. I am only proving that there is one real root, not that there is only one real root.
The derivative of the function f(x) = 2x^3+12x^2+18x+24 is given by f'(x) = 6x^2 + 24x + 18
f'(x) = 6*(x^2 + 4x + 3)
Let's take two values of x, x = 0 and x = -2.
At x = 0
f'(x) = 18 which is positive
and for x = -2
f'(x) = -6 which is negative
By the Intermediate Value Theorem there exists a number "c" between -2 and 0 such that f(c) = 0. This shows that the equation has a real root.
This proves that 2x^3+12x^2+18x+24 = 0 has one real root.
First, we'll simplify the given equation, by dividing by 2:
x^3 + 6x^2 + 9x + 12 = 0
To determine the number of real roots of the equation, we'll have to create the Rolle's string. According to Rolle's string, between 2 consecutive roots of derivative, we'll find a real root of the equation, if and only if the product of the values of derivatives, is negative.
Since only a continuous function could be differentiated, we'll check the continuity of the function. The Rolle's string could be applied if and only if the polynomial function is continuous.
We'll evaluate the limits of the function, if x approaches to + and - infinite.
lim f(x) = lim (x^3+6x^2+9x+12) = + infinite, for x approaches to +infinite.
To determine the Rolle's string we need to determine the roots of the 1st derivative of the function.
f'(x) = (x^3+6x^2+9x+12)'
f'(x) = 3x^2 + 12x + 9
We'll put f'(x) = 0
3x^2 + 12x + 9 = 0
We'll divide by 3:
x^2 + 4x + 3 = 0
x1 = -1 and x = -3
Now, we'll calculate the values of the function for each value of the roots of the derivative.
f(-inf.) = lim f(x) = -inf
f(+inf.) = lim f(x) = +inf.
f(-1) = -1+6-9+12=8
f(-3) = -27 + 54 - 27 + 12 = 12
The values of the function represents the Rolle's string.
-inf. 12 8 +inf.
We notice that the sign varies 1 time, between -infinite and -3:
Therefore, the equation will have one real root that belongs to the range: (-infinite ; -3).
We’ve answered 319,195 questions. We can answer yours, too.Ask a question