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

PLEASE HELPPP MEEEEEEEEEE

qwelly [4]

Answer:

x = 17°

Step-by-step explanation:

<u>Angles And Parallel Lines </u>

If two parallel lines are crossed by a third line, corresponding angles are congruent. Also, consecutive angles are supplementary

We can clearly see in the image provided, being w and b parallel, that the corresponding angles (4x-3) and 65° are congruent. We can express that relation as

$4x-3=65$

Operating

$4x=65+3=68$

$x=\frac{68}{4}=17$

So the value of the angle x is 17°

8 0

3 years ago