Please help Me. NO LINKS PLEASE. 8th Grade
1 answer:
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By definition of absolute value, you have
or more simply,
On their own, each piece is differentiable over their respective domains, except at the point where they split off.
For <em>x</em> > -1, we have
(<em>x</em> + 1)<em>'</em> = 1
while for <em>x</em> < -1,
(-<em>x</em> - 1)<em>'</em> = -1
More concisely,
Note the strict inequalities in the definition of <em>f '(x)</em>.
In order for <em>f(x)</em> to be differentiable at <em>x</em> = -1, the derivative <em>f '(x)</em> must be continuous at <em>x</em> = -1. But this is not the case, because the limits from either side of <em>x</em> = -1 for the derivative do not match:
All this to say that <em>f(x)</em> is differentiable everywhere on its domain, <em>except</em> at the point <em>x</em> = -1.
Answer:
Options (A) and (C)
Step-by-step explanation:
From table attached,
Let the equation of a linear equation that represents the input-output values in the table is,
y - y' = m(x - x')
Where (x', y') is a point lying on the graph.
And m = slope of the line passing through two points and
m =
Slope of the line passing through (0, 11) and (1, 5) will be,
m =
m = -6
Therefore, equation of the line passing through (0, 11) and slope = -6 will be,
y - 11 = -6(x - 0)
Equation of a line passing through (1, 5) and slope = -6 will be,
y - 5 = -6(x - 1)
Equation of a line passing through (2, -1) and slope = -6 will be,
y + 1 = -6(x - 2)
Equation of a line passing through (3, -7) and slope = -6 will be,
y + 7 = -6(x - 3)
Therefore, Options (A) and (C) are the correct options.