Answer:
24 Domain: s>=2 or s<=-2
25. 3x^2 +14x +10
26. x^2 -2x+5
Step-by-step explanation:
24. Domain is the input or s values
square roots must be greater than or equal to zero
s^2-4 >=0
Add 4 to each side
s^2 >=4
Take the square root
s>=2 or s<=-2
25. f(g(x)) stick g(x) into f(u) every place you see a u
f(u) = 3u^2 +2u-6
g(x) = x+2
f(g(x) = 3(x+2)^2 +2(x+2) -6
Foil the squared term
= 3(x^2 +4x+4) +2x+4-6
Distribute
= 3x^2 +12x+12 +2x+4-6
Combine like terms
=3x^2 +14x +10
26 f(g(x)) stick g(x) into f(u) every place you see a u
f(u) = u^2+4
g(x) = x-1
f(g(x) = (x-1)^2 +4
Foil the squared term
= (x^2 -2x+1) +4
= x^2 -2x+5
Answer:
99.66%
Step-by-step explanation:
The vertex (minimum) of the quadratic ax² +bx +c is located at x=-b/(2a). This means the minimum value of f(x) will be found at x = -3/(2*1) = -1.5.
Since the vertex of the quadratic is less than 0, the maximum value of the quadratic will be found at x=2, the end of the interval farthest from the vertex.
On the given interval, ...
the absolute minimum value of f is f(-1.5) = ln(1.75) ≈ 0.559616
the absolute maximum value of f is f(2) = ln(14) ≈ 2.639057
They are equal its just rounded up to one decimal place