What is the average rate of change of f(x) = -x2 + 3x + 6 over the interval –3

Mathematics

1 answer:

Rufina [12.5K]3 years ago

3 0

Answer:

$\frac{\Delta y}{\Delta x} =\frac{f(x_2)-f(x_1)}{x_2-x_1} =\frac{6-(-12)}{3-(-3)} =\frac{18}{6}= 3$

Step-by-step explanation:

To find the average rate of change of a function over a given interval, basically you need to find the slope. The mathematical definition of the slope is very similar to the one we use every day. In mathematics, the slope is the relationship between the vertical and horizontal changes between two points on a surface or a line. In this sense, the slope can be found using the following expression:

$Average\hspace{3}rate\hspace{3}of\hspace{3}change=Slope=\frac{y_2-y_1}{x_2-x_1} =\frac{f(x_2)-f(x_1)}{x_2-x_1}$

So, the average rate of change of:

$f(x)=-x^2+3x+6$

Over the interval $-3$

Is:

$f(x_2)=f(3)=-(3)^2+3(3)+6=-9+9+6=6\\\\f(x_1)=f(-3)=-(-3)^2+3(-3)+6=-9-9+6=-12$

$\frac{\Delta y}{\Delta x} =\frac{f(x_2)-f(x_1)}{x_2-x_1} =\frac{6-(-12)}{3-(-3)} =\frac{18}{6}= 3$

Therefore, the average rate of change of this function over that interval is 3.

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A projectile is fired with muzzle speed 250 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe

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where $g=9.8\dfrac{\rm m}{\mathrm s^2}$ . Solve $y=0$ for the time $t$ it takes for the projectile to reach the ground:

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$v_x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ$

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At the time the projectile hits the ground, its velocity vector has horizontal component approx. 176.77 m/s and vertical component approx. -178.43 m/s, which corresponds to a speed of about $\sqrt{176.77^2+(-178.43)^2}\dfrac{\rm m}{\rm s}\approx250\dfrac{\rm m}{\rm s}$ .