## Thomas' Calculus 13th Edition

$(f\circ g\circ h)(x)=6x^2+1.$
We are given $f(x)=3x+4$, $g(x)=2x-1$, and $h(x)=x^2$. So: $$(f\circ g\circ h)(x)=f(g(h(x)))=f(g(x^2))=f(2x^2-1)=3(2x^2-1)+4\\=6x^2-3+4 =6x^2+1.$$