Answer:
f(0) = 1/2
Step-by-step explanation:
At x=0, the inequality tells you ...
1/2 ≤ 1 -f(0) ≤ 1/2
That is, ...
1 - f(0) = 1/2
f(0) = 1/2

Answer:
The translation is right 3 down 2 which is reflection
Step-by-step explanation:
Step-by-step explanation:
mark me as brilliant plzzz
Answer:
b = 1, c = -1 and d = 4
Step-by-step explanation:
To solve this question the rule of multiplicity of a polynomial is to be followed.
If the multiplicity of a polynomial is even at a point, graph of the polynomial will touch the x-axis.
If the multiplicity of the polynomial is odd, graph will cross the x-axis at that point.
From the graph of function 'f',
f(x) = (x - b)(x - c)²(x - d)³
Since, graph of the function 'f' crosses x-axis at x = 1 and x = 4, multiplicity will be odd and touches the x-axis at x = -1 multiplicity will be even.
So the function will be,
f(x) = (x - 1)[x - (-1)]²(x - 4)³
Therefore, b = 1, c = -1 and d = 4 will be the answer.
The scenario can be described using a piecewise function like:
f(x) = 1/x if x < c.
f(x) = x if x = c
f(x) = 1/(x + 73) if x > c.
<h3>
When the value exists but the limit does not?</h3>
Remember that the limit only exists if the limit from left and the limit from the right give the same value.
Then, we can just define a piecewise function of the form:
f(x) = 1/x if x < c.
f(x) = x if x = c
f(x) = 1/(x + 73) if x > c.
Clearly, this is not a continuous function.
Notice that:

So the limits from left and right are different, then:

Does not exist.
If you want to learn more about limits:
brainly.com/question/5313449
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