= Solution
Write $\phi(\alpha)=f(x+\alpha s)$ and $h(\alpha)=\phi(\alpha)-\phi(0)-\beta_1\alpha d$. Part (a) gives $h(\alpha)<0$ just to the right of zero. If $L$ is a lower bound for $f$, then
$$
h(\alpha)\geq L-\phi(0)-\beta_1\alpha d\longrightarrow+\infty,
$$
because $d<0$. By <continuity>, there is a first positive zero $A$ of $h$, and $h(\alpha)\leq0$ on $[0,A]$. Apply the <mean value theorem> to $\phi$ on this interval: for some $t\in(0,A)$,
$$
\nabla f(x+ts)^Ts=\phi'(t)=\frac{\phi(A)-\phi(0)}A=\beta_1d>\beta_2d.
$$
At this same $t$, $h(t)\leq0$ supplies the <Armijo condition>. Thus $\alpha=t>0$ satisfies both weak <Wolfe conditions>. The boundedness assumption prevents the <objective function> from following its initial negative slope forever; this is the mechanism behind <Existence of a Wolfe step>.
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