Find the coefficient of the squared term in the simplified form for the second derivative., f"(x) for f(x)=(x^3+2x+3)(3x^3-6x^2-8x+1). use the hyphen for negative values.
( fill-in-the-blank)

Use the product rule:
f'(x) =(x³+2x+3)'(3x³-6x²-8x+1) + (x³+2x+3)(3x³-6x²-8x+1)'
f'(x) =(3x²+2)(3x³-6x²-8x+1) + (x³+2x+3)(6x²-12x-8)

Use the product rule again:
f''(x) = (6x)(3x³-6x²-8x+1) + (3x²+2)(9x²-12x-8) + (3x²+2)(6x²-12x-8) + (x³+2x+3)(12x-12)

We only care about the coefficient of the x² term so let's extract the operations of terms that give us x²: (6x)(-8x)+(3x²)(-8)+2(9x²)+(3x²)(-8)+2(6x²)+(2x)(12x) = - 48x²+(-24x²)+18x²+(-24x²)+12x²+24x² = 54x²

    y = −1.5x + 2.5 Step-by-step explanation:

       For us to write the equation for this line, we need to (1) find the slope of the line, and (2) use one of the points to write an equation:  

       The question gives us two points (one of which is the y-intercept), (4.5,-4.25), and (0, 2.5), from which we can find the slope and later the equation of the line.

 

Finding the Slope    

The slope of the line (m) = (y₂ - y₁) ÷ (x₂ - x₁)    

                                        =  (2.5 - (- 4.25)) ÷ (0 - 4.5)

                                        =  −1.5

                                   

 Finding the Equation

Now that we know the slope and the y-intercept, we can write the equation of the line in the point-slope form (y = mx + b)

                                            y = −1.5x + 2.5

Checking my

I used Desmos to graph the line I got and the points. Since the line passes through the points then the line is correct.