Question

The interval on which the function \(f(x)=2x^3 +12x^2 +18x-7\) is decreasing, is:

• A. [-3,-1]
• B. (-2,-1)
• C. [-3,0]
• D. [-2,-1]

Answer 1

The Correct Option is A

To determine the interval on which the function \(f(x)=2x^3+12x^2+18x-7\) is decreasing, we need to find where the derivative of the function, \(f'(x)\), is negative. Begin by calculating the derivative: \(f'(x)=\frac{d}{dx}(2x^3+12x^2+18x-7)=6x^2+24x+18\).
Next, find the critical points by setting \(f'(x)\) to zero: \(6x^2+24x+18=0\).
Divide the entire equation by 6 to simplify: \(x^2+4x+3=0\).
Factor the quadratic equation: \((x+1)(x+3)=0\).
This gives the critical points \(x=-1\) and \(x=-3\).
Now, determine the sign of \(f'(x)\) in the intervals created by these critical points: \((-∞,-3)\), \((-3,-1)\), and \((-1,∞)\).
Pick test points in each interval:

• For \((-∞,-3)\), choose \(x=-4\): \(f'(-4)=6(-4)^2+24(-4)+18=54\), positive.
• For \((-3,-1)\), choose \(x=-2\): \(f'(-2)=6(-2)^2+24(-2)+18=-6\), negative.
• For \((-1,∞)\), choose \(x=0\): \(f'(0)=6(0)^2+24(0)+18=18\), positive.

Thus, \(f(x)\) is decreasing in the interval \([-3,-1]\) where \(f'(x)<0\). Therefore, the correct answer is the interval \([-3,-1]\).