Confused

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) .