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. 

Julius  Feb 27, 2018

1+0 Answers


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) .

hectictar  Feb 28, 2018

30 Online Users

New Privacy Policy (May 2018)
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  Privacy Policy