Answer:
2a : f(-2) = 15
2b : f(2x) = 8x² - 2x + 5
2c : f(x-1) = 2x² - 5x + 8
3a : x > 1/3
3b : x <> {2,3}
4 : 6x
Step-by-step explanation:
2a.
2×(-2)² - (-2) + 5 = 2×4 + 2 + 5 = 15
2b.
2×(2x)² - 2x + 5 = 2×4x² - 2x + 5 = 8x² - 2x + 5
2c.
2×(x-1)² - (x-1) + 5 = 2×(x² - 2x + 1) - x + 1 + 5 =
= 2x² - 4x + 2 - x + 1 + 5 = 2x² - 5x + 8
3a.
the square root is not defined for negative values, which would happen for any x < 1/3. and for x=1/3 the expression would be a division by 0, which is also undefined.
so, all that is left is x>1/3.
3b.
we must avoid a division by 0. everything else is well defined.
so, when is that expression x² - 5x + 6 = 0 ?
solution for a quadratic equation is
x = (-b ± sqrt(b² - 4ac))/2a
a = 1
b = -5
c = 6
x = (5 ± sqrt(25 - 24))/2 = (5 ± sqrt(1))/2 = (5 ± 1)/2
x1 = (5+1)/2 = 6/2 = 3
x2 = (5-1)/2 = 4/2 = 2
so, these values must be avoided for x. everything else is valid.
4.
(f(x+h) - f(x))/h
for h going to 0.
f(x) = 3x² - 5
f(x+h) = 3×(x+h)² - 5
f(x+h) - f(x) = 3×(x² + 2hx + h²) -5 - 3x² + 5 =
= 3x² + 6hx + 3h² - 3x² = 6hx + 3h²
=>
(f(x+h) - f(x))/h = (6hx + 3h²)/h = 6x + 3h = for h 0 = 6x
