EXAMPLE 5 Suppose that f(0) = −4 and f '(x) ≤ 5 for all values of x. How large can f(2) possibly be? SOLUTION We are given that f is differentiable (and therefore continuous) everywhere. In particular, we can apply the Mean Value Theorem on the interval [0, 2] . There exists a number c such that f(2) − f(0) = f '(c) − 0 so f(2) = f(0) + f '(c) = −4 + f '(c). We are given that f '(x) ≤ 5 for all x, so in particular we know that f '(c) ≤ . Multiplying both sides of this inequality by 2, we have 2f '(c) ≤ , so f(2) = −4 + f '(c) ≤ −4 + = . The largest possible value for f(2) is .

step-by-step explanation:

solution:

we have to construct

∠m nt ≅ ∠p qr

the steps to draw this construction are

step 1: use a compass to draw an arc from point q which intersects the side pq at point a and the side qr at point b.

step 2: draw a segment nt and use the same width of the compass to draw an arc from point n which intersects the segment nt at a point x.

step 3:

step 4: join points n and y using a straightedge.

step 3:

option (d).maintaining the same width of the compass as ab, draw an arc from point x such that it intersects the arc drawn from n in a point y.

to draw   congruent angles , 1. width of compasses while drawing two angles should be same i.e from point at which angle has to be drawn, 2.the distance between the two points where the compass intersects the two sides should be same while drawing any two congruent angles.

answer: 819.89 miles across america

step-by-step explanation: