+956 
My best shot at this was to find the derivative of f(x), but then how do we find the derivative of b? Do we assume it's a constant? Kinda confused on this.
+9479 Determine the value of b if the slope of the tangent to f(x) = (3x2 + 1)(2x2 + b) at x = -1 is -16 .
\(f(x)\,=\,(3x^2+1)(2x^2+b) \\~\\ f(x)\,=\,6x^4+3bx^2+2x^2+b \\~\\ \frac{d}{dx}f(x)\,=\,\frac{d}{dx}(6x^4+3bx^2+2x^2+b) \\~\\ f'(x)\,=\,\frac{d}{dx}(6x^4+3bx^2+2x^2+b) \\~\\ f'(x)\,=\,\frac{d}{dx}6x^4+\frac{d}{dx}3bx^2+\frac{d}{dx}2x^2+\frac{d}{dx}b \)
We are trying to find the value of b , so it must be a constant. As the value of x changes, the value of b doesn't change. So \(\frac{d}{dx}b=0\)
\(f'(x)\,=\,24x^3+6bx+4x+0 \\~\\ f'(x)\,=\,24x^3+6bx+4x\)
The slope at x = -1 is -16 . That means f'(-1) = -16
\(f'(-1)\,=\,24(-1)^3+6b(-1)+4(-1) \\~\\ -16\,=\,24(-1)^3+6b(-1)+4(-1) \\~\\ -16\,=\,-24-6b-4 \\~\\ -16\,=\,-28-6b \\~\\ 12\,=\,-6b \\~\\ b=-2\)
To check this, here's a graph of f(x) and a line that passes through f(x) at the point where x = -1 with a slope of -16 . We can see that the line appears tangent to f(x) .
