Answer:
The last equation x2 - 2x -4 = 0
has solution (x - 1)^2 - 5 = 0, x = 1 + root(5) or x = 1 - root(5)
Step-by-step explanation:
If a quadratic function has roots 1 and 5
f(x) = (x -1)(x- 5)
f(x) = x^2 - 6x + 5
Unless you meant. -4 and 6 ?
g(x) = (x + 4)(x - 6)
g(x) = x^2 -2x -24
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Or did you mean x = 1 and x =4 ?...
x^2 + 2x + 4 = 0 : complete square x^2 + 2x + 1 + 3 = 0, (x+1)^2 + 3 = 0
x^2 - 2x + 4 = 0 : complete square: (x -1)^2 + 3 = 0
0x^2 + 2x - 4 = 0, 2x - 4 = 0, x = 2
x^2 - 2x - 4 = 0 becomes: x^2 - 2x + 1 - 1 -4 = 0 ; (x - 1)^2 - 5 = 0
To check if a piecewise defined function is continuous, you need to check how the pieces "glue" together when you step from one domain to the other.
So, the question is: what happens at x=3? If you reach x=3 from values slightly smaller than 3, you obey the rule f(x)=log(3x). So, as you approach 3, you get values closer and closer to

Similarly, if you reach x=3 from values slightly greater than 3, you obey the rule f(x)=(4-x)log(9). So, as you approach 3, you get values closer and closer to

So, the function is continuous at x=3, because both pieces approach log(9) as x approaches 3.
-3x, -x + 1/2, 2x, 10x if you need anything else feel free to ask. I would like brainliest but if not its ok
Answer:
6v+10
Step-by-step explanation:
hope this helped
can also write 6*v+10
F(3)=-4(3)+3=-9
g(3)=3(3)+4=13
f(3).g(3)=-9*13=-117