A function $f$ has domain $ ext{R}$ and range $ ext{y} ext{ } ext{e}: ext{y} ext{ } ext{≥} ext{e}$ The graph of $y = f(x)$ is shown - AQA - A-Level Maths Pure - Question 7 - 2018 - Paper 2

Now applying integration by parts:

$\int u \, dv = uv - \int v \, du$

Substituting our values:

$f(x) = (x - 1)e^x - \int e^x \, dx$

The integral of $e^x$ is simply $e^x$, thus:

$f(x) = (x - 1)e^x - e^x + C$

Combining the terms gives:

$f(x) = (x - 2)e^x + C$

where $C$ is the constant of integration.

To find the constant $C$, observe the minimum $y$ value in reference to the range of the function. The range states that $y ext{ } ≥ ext{ } e$.

The minimum occurs when rac{dy}{dx} = 0:

$0 = (x - 1)e^x$ So, $x = 1$ is the point of interest. At this point:

$f(1) = (1 - 2)e^1 + C = -e + C$ Setting this equal to the minimum value at $y = e$ gives:

$-e + C = e$ Thus, $C = 2e$.

Substituting back the value of $C$ into our expression for $f(x)$:

$f(x) = (x - 2)e^x + 2e$

This is the required final expression for $f(x)$.