Question

If dy/dx=20/(x+1)^3
Find d^2y/dx^2 and the real values of x for which d^2y/dx^2 = -15/4

Answer 1

\[\frac{dy}{dx}=\frac{20}{(x+1)^3}=20(x+1)^{-3}\\
\frac{d^2y}{dx^2}=-60(x+1)^{-4}=-\frac{60}{(x+1)^4}\]
Now you solve for x in the following equation:
\[-\frac{60}{(x+1)^4}=-\frac{15}{4}\]

Answer 2

So you rearrange \[-\frac{ 60 }{(x+1)^4 } =-\frac{ 15 }{ 4 }\] with respect to x.
does that give \[x=\sqrt[4]{\frac{ 15 }{ 4\times-60 }}-1\] ?
if so I get a math error

Answer 3

\[\begin{align*}
-\frac{60}{(x+1)^4}&=-\frac{15}{4}\\
\frac{60}{(x+1)^4}&=\frac{15}{4}\\
\frac{4}{(x+1)^4}&=\frac{1}{4}\\
\frac{16}{(x+1)^4}&=1\\
16&=(x+1)^4\\
\pm\sqrt[4]{16}&=x+1\\
\pm2&=x+1\\
x=1~&~x=-3
\end{align*}\]

Answer 4

x is equal to 1 or -3

Answer 5

Ah ok my mistake I read value of x instead of values, thanks for the help