Question

Let $F\left(x\right)=f\left(x\right)g\left(x\right)h\left(x\right)$ for all real x, where$f\left(x\right),g\left(x\right)$and$h\left(x\right)$ are differentiable functions. At some point $x_0$, $F^{\prime }\left(x_0\right)=21F\left(x_0\right)$, $f^{\prime }\left(x_0\right)=4f\left(x_0\right)$, $g^{\prime }\left(x_0\right)=-7g\left(x_0\right)$ and $h^{\prime }\left(x_0\right)=kh\left(x_0\right)$. Then $k$ is equal to

Answer 1

$F^{\prime }\left(x\right)=f^{\prime }\left(x\right)g\left(x\right)h\left(x\right)+f\left(x\right)g^{\prime }\left(x\right)h\left(x\right)+f\left(x\right)g\left(x\right)h^{\prime }\left(x\right)$
so ,$21F\left(x_0\right)=F^{\prime }\left(x_0\right)=f^{\prime }\left(x_0\right)g\left(x_0\right)h\left(x_0\right)+f\left(x_0\right)g^{\prime }\left(x_0\right)h\left(x_0\right)+f\left(x_0\right)g\left(x_0\right)h^{\prime }\left(x_0\right)$
$=4f\left(x_0\right)g\left(x_0\right)h\left(x_0\right)-7f\left(x_0\right)g\left(x_0\right)h\left(x_0\right)+kf\left(x_0\right)g\left(x_0\right)h\left(x_0\right)$
$=(-3+k)f\left(x_0\right)g\left(x_0\right)h\left(x_0\right)$
$=(-3+k)F\left(x_0\right)$
Hence $21=-3+k\Rightarrow k=24$