For the following exercises, evaluate the function at the indicated values: f (−3); f (2); f (−a); −f (a); f (a + h). 5. f (x) = −2x 2 + 3x
6. f (x) = 2|3x − 1|

5. f(x) = -2x² + 3x  

f(-3) = -2(-3)² + 3(-3) = -27

f(2) = -2(2)² + 3(2) = -2

f(-a) = -2(-a)² + 3(-a) = -2a² - 3a

-f(a) = -[-2a² + 3a] = 2a² - 3a

f(a + h) = -2(a + h)² + 3(a + h) = -2(a² + 2ah + h²) + 3a + 3h = -2a² - 4ah - 2h² + 3a + 3h

6. f(x) = 2|3x - 1|

f(-3) = 2|3(-3) - 1| = 2*10 = 20

f(2) = 2|3(2) - 1| = 2*5 = 10

f(-a) = 2|3(-a) - 1| = 2|-3a - 1|

-f(a) = -(2|3a - 1|) = -2|3a - 1|

f(a + h) = 2|3(a + h) - 1| = 2|3a + 3h - 1|

answer: it's d

step-by-step explanation:

explanation:

infrared radiation has wavelength in the range between 700 nm and 1 mm. lets' convert these values into scientific notation, keeping in mind that:

let's start from the lower value, we have:

(we moved the dot two places to the left, so we increased the exponent by 2 units, from -9 to -7)

the higher value is:

so, the range of wavelength of infrared radiation is: