1. The problem asks whether the term "grow" in the phrase "f(t) must not grow faster than a function of exponential type" refers to convergence in the context of the Laplace transform. 2. The Laplace transform is defined as $$L\{f(t)\} = \int_0^\infty e^{-st} f(t) \, dt$$ where the integral is improper because the upper limit is infinity. 3. For the integral to converge, the product $$e^{-st} f(t)$$ must approach zero sufficiently fast as $$t \to \infty$$. 4. The phrase "f(t) must not grow faster than a function of exponential type" means that $$f(t)$$ cannot increase faster than some exponential function like $$e^{ct}$$ for some constant $$c$$. 5. This condition ensures that $$e^{-st}$$ dominates $$f(t)$$ for sufficiently large $$t$$, making the integral converge. 6. Therefore, "grow" here refers to the growth rate of $$f(t)$$ and is directly related to the convergence of the Laplace integral. 7. In summary, yes, "grow" in this context refers to the behavior of $$f(t)$$ that affects the convergence of the Laplace transform integral. Final answer: The term "grow" refers to the growth rate of $$f(t)$$ and its relation to the convergence of the Laplace transform integral.