Recover a function from its gradient and one value

Strict increase requires $f^{\prime}>0$; concavity downwards requires $f^{\prime\prime}<0$, so $f^{\prime}$ must be decreasing. Both conditions hold on the right-hand branch, $x>1$.

The zeros of $f^{\prime}$ are $x=-3$ and $x=0$.

At $-3$, the derivative touches zero and changes from increasing to decreasing, so $f$ changes concavity without turning. At $0$, its gradient changes from negative to positive.

The minimum is $(0,1)$, which is the y-intercept. The graph decreases up to this minimum, with a stationary inflection at $x=-3$ and another change of concavity at $x=-1.5$; it then increases for $0<x<1$. It stays at or above $1$ for $x<1$, so has no x-intercepts. The source labels the two negative x-values but does not assign their function values.

The printed answer draws $x=1$ as a vertical asymptote of $f$, with the right-hand end tending to $+\infty$. The SVG retains that printed qualitative sketch.

The gradient graph alone does not guarantee that $f$ diverges at $x=1$: a derivative can tend to $+\infty$ while having a finite improper integral. Thus the vertical asymptote in the printed answer is an additional smooth-shape interpretation; it requires a non-integrable gradient near $1$. The stationary points, sign intervals, y-intercept and absence of x-intercepts above follow directly from the supplied gradient and $f(0)=1$.