Answer:
1) [tex]\frac{dy}{dx}=12x^2+6x-6[/tex]
2) [tex]x\leq-1\text{ or } x\geq 1/2[/tex]
Step-by-step explanation:
So we have the equation:
[tex]y=4x^3+3x^2-6x-1[/tex]
Part A)
We want to find the derivative of the above function. So:
[tex]\frac{dy}{dx}=\frac{d}{dx}[4x^3+3x^2-6x-1][/tex]
Expand:
[tex]\frac{dy}{dx}=\frac{d}{dx}[4x^3]+\frac{d}{dx}[3x^2]+\frac{d}{dx}[-6x]+\frac{d}{dx}[-1][/tex]
Use the power rule. Therefore, our derivative is:
[tex]\frac{dy}{dx}=3(4x^2)+2(3x)+(-6)+(0)[/tex]
Simplify:
[tex]\frac{dy}{dx}=12x^2+6x-6[/tex]
Part B)
We want to find the range of the values of x such that:
[tex]\frac{dy}{dx}\geq 0[/tex]
So, all we have to do is substitute our derivative in:
[tex]12x^2+6x-6\geq0[/tex]
And solve for this quadratic inequality!
Solve it normally. Pretend the inequality is with an equal sign. So:
[tex]12x^2+6x-6=0[/tex]
Divide everything by 6:
[tex]2x^2+x-1=0[/tex]
Factor. We can use -2 and 1. So:
[tex]2x^2+2x-x-1=0[/tex]
From the first two terms, factor out 2x.
From the third and fourth terms, factor out a negative. So:
[tex]2x(x+1)-(x+1)=0[/tex]
Grouping:
[tex](2x-1)(x+1)=0[/tex]
Zero Product Property:
[tex]2x-1=0\text{ or } x+1=0[/tex]
Solve for x in each case:
[tex]x=1/2\text{ or } x=-1[/tex]
So, the zeros of our function is -1 and 1/2.
Looking above, we can see that this is a greater than inequality.
Since this is a greater than inequality, our solution is all values to the left of the first zero and all values to the right of the second zero.
Therefore, our solution is:
[tex]x\leq-1\text{ or } x\geq 1/2[/tex]
In interval notation, this is:
[tex](-\infty,-1]\cup[1/2,\infty)[/tex]
So, for all values within the above interval, our derivative, dy/dx, will be greater than or equal to 0.
And we're done!
Edit: Typo. Brainly is rather buggy right now.
