Calculate maximum of g(x)

We have that $g(x) = \sqrt{13} \left(\dfrac{2}{\sqrt{13}}\cos(x) + \dfrac{3}{\sqrt{13}}\sin(x)\right) = \sqrt{13} \sin(x + \theta)$ , where $\theta = \arcsin \left(\dfrac{2}{\sqrt{13}}\right)$ .

Then since the range of the sine function is [ $-1,1$ ] we see that the maximum value of $g(x)$ is $\sqrt{13} = \boxed{3.6}$ to one decimal place.

Let us investigate dg(x)/dx = -2*Sin(X) + 3 * Cos(X). To find out the maximum point and maximum value dg(x)/dx = 0 implies that Tan(x) = 1.5

X = arctan 1.5 = tan⁻¹ 1.5 = 56° 18' + 2*N * PI

Let us investigate the second derivative = -2 *Cos(X) - 3 * Sin (X), it is less than 0 at X = 56° 18'

Maximum value of g(x) at X = 56° 18'

2*0.55 + 3 * 0.83 = 3.6