Sketch the graphs of $y=3e^{x}$ and $y=x+3$ on the same diagram. Indicate clearly the coordinates of the points of intersection between the 2 graphs. Solve the inequality $3e^{x} > x + 3$. (b) Hence find $\displaystyle \int_{-2}^{2} |2e^{x} - x - 3|\, dx$, giving your answer in an exact form.

1. Sketch the graphs of $y=3e^x$ and $y=x+3$ on the same axes.

• Sketch the straight line $y = x + 3$ passing through the intercepts $(-3, 0)$ and $(0, 3)$.
• Sketch the exponential curve $y = 3e^x$, which is strictly increasing, has a horizontal asymptote at the negative $x$-axis ($y = 0$), and passes through $(0, 3)$.
• Ensure the relative positions show two points of intersection (one at the $y$-axis and one in the second quadrant).

2. Find the coordinates of the points of intersection.

Equate the functions to find the intersection points:
$3e^x = x + 3$
Using a graphing calculator, the $x$-coordinates are exactly $x = 0$ and approximately $x \approx -2.8214$.
For $x = 0$, $y = 3(e^0) = 3$.
For $x \approx -2.8214$, $y = x + 3 \approx -2.8214 + 3 = 0.1786$.
Rounding to 3 significant figures, the coordinates of the points of intersection are $(0, 3)$ and $(-2.82, 0.179)$.

3. Solve the inequality $3e^x > x + 3$.

The inequality $3e^x > x + 3$ requires finding the range of $x$ where the curve $y = 3e^x$ lies strictly above the line $y = x + 3$.
By observing the sketched graph and the intersection points, the curve is above the line for $x$-values to the left of the negative intersection point and to the right of the $y$-axis.
Therefore, the solution is $x < -2.82$ or $x > 0$.

1. Use part (a) to split the modulus

From part (a), $3e^{x}>x+3$ (i.e. $3e^{x}-x-3>0$) for $x>0$, and the only other root lies below $x=-2$, so $3e^{x}-x-3<0$ for $-2\le x<0$.

Hence
$\int_{-2}^{2}\big|3e^{x}-x-3\big|\,dx=\int_{-2}^{0}-(3e^{x}-x-3)\,dx+\int_{0}^{2}(3e^{x}-x-3)\,dx.$

2. Evaluate the two integrals and combine

Let $F(x)=3e^{x}-\tfrac{x^{2}}{2}-3x$.

$\int_{-2}^{0}-(3e^{x}-x-3)\,dx=F(-2)-F(0)=(3e^{-2}+4)-3=3e^{-2}+1.$

$\int_{0}^{2}(3e^{x}-x-3)\,dx=F(2)-F(0)=(3e^{2}-8)-3=3e^{2}-11.$

Total $=3e^{-2}+1+3e^{2}-11=3e^{2}+3e^{-2}-10.$