<s>[INST] Calculus differentiable gra Hey guys so stuck on a calculus question. So far all I know is that $d$ should equal $4$. I then got the derivative of the a b c d function to $3ax^2 + 2bx + c$ , subbed in $(0,4)$ to get $c$, which was also $4$. Just wondering if I'm on the right track and where I should go from here.  

A section of road, represented by the line $y = x + 4$ when $x \leq 0$, is to
  be smoothly connected to another section of road, represented by $y = 4
> –x$ when $x \geq 4$ , by means of a curved section of road, represented by a
  cubic curve $y = ax^3+ bx^2+ cx + d$ . Find $a, b, c$ and $d$ such that the
  function $f(x)$ is everywhere differentiable (and therefore everywhere
  continuous), where
$$f(x) = \begin{cases}
x+4 &x\leq 0 \\ 
ax^3 +bx^2 +cx +d  &0<x<4 \\ 
 4-x &x\geq 4 
\end{cases}$$

 [/INST] ok, so 
$y = $
$x + 4, x ≤0$
$ax^3+ bx^2+ cx + d, 0 < x < 4$
$4 – x, x ≥4$
and y has to be differentiable everywhere (and therefor continuous, too)
so, the derivative at x = 0 and at x = 4 have to be the same when approached from both sides, as do the values of y. 
Let $P(x) = ax^3 + bx^2 + cx + d$
and
$DP(x) = 3ax^2 + 2bx + c$ (the derivative of P)
so, the limit as $x$ goes to $0$ of $P = f(0) = 4$, so $P(0) = 4$, so $d = 4$.
$DP(0) = 1$, and $DP(4) = -1$ (in order for it to be differentiable)
$DP(0) = c = 1 $
so $c=1$, $d=4$
$DP(4) = -1 = 3a*16 + 2b*4 + 1$
$2 + 48a + 8b = 0 = 1 + 24a + 4b$
and finally, $P(4) = 4 - 4 = 64a + 16b + 4(1) + 4$
$0 = 64a + 16b + 8$
$0 = 16a + 4b + 2$
so
$24a + 4b + 1 = 16a + 4b + 2$
$24a - 16a = 1$
$a = 1/8$
then 
$0 = 16(1/8) + 4b + 2$
$0 = 2 + 2 + 4b$
$-4 = 4b$
$b = -1$
In conclusion: 
$a=1/8$, $b=-1$, $c=1$, $d=4$
hope it helps. (sorry for the for the poor formatting.)
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