The maxima of f(x) occur at its critical points, where f '(x) is zero or undefined. We're given f '(x) is continuous, so we only care about the first case. Looking at the plot, we see that f '(x) = 0 when x = -4, x = 0, and x = 5.

Notice that f '(x) ≥ 0 for all x in the interval [0, 5]. This means f(x) is strictly increasing, and so the absolute maximum of f(x) over [0, 5] occurs at x = 5.

By the fundamental theorem of calculus,

$\displaystyle f(5) = f(0) + \int_0^5 f'(x) \, dx$

The definite integral corresponds to the area of a trapezoid with height 2 and "bases" of length 5 and 2, so

$\displaystyle \int_0^5 f'(x) \, dx = \frac{5+2}2 \times 2 = 7$

$\implies \max\{f(x) \mid 0\le x \le5\} = f(5) = f(0) + 7 = \boxed{13}$

8 0

2 years ago

Do I solve this by equaling both equations to each other?

Mars2501 [29]

Yes, you got it! You would need to set the equations equal to each other, because that represents when their populations are the same. Then, solving for x will find the time in months until you reach that equilibrium in population. If you need help solving for x, just comment on this. Hope it helps!

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%5Cfrac%7B3%7D%7B16%7D%20%7D%20" id="TexFormula1" title=" \sqrt{ \frac{3}{16}

sweet-ann [11.9K]

Hello!
When you have a non-exact root (irrational number)

$\sqrt{ \frac{3}{16} } \to \frac{ \sqrt{3} }{ \sqrt{16} } \to \boxed{\boxed{\frac{ \sqrt{3} }{4} }}\Longleftarrow(irrational\:number)\end{array}}\qquad\quad\checkmark$

7 0

3 years ago