Answer:
- f'(1) exists: f'(1) = -2
- f'(0) DNE
Step-by-step explanation:
<h3>a)</h3>
The function is continuous for x > 0 . The derivative is defined on that interval and is equal to ...
f'(x) = -2x . . . . . for x > 0
Then at x = 1, the derivative is ...
f'(1) = -2(1)
f'(1) = -2
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<h3>b)</h3>
The function has a jump discontinuity at x=0, so the derivative does not exist at that point. A condition for the existence of the derivative is that the function is continuous at the point of interest.
Answer:
B) y = (-7X/5) + 4
Step-by-step explanation:
Recall two lines are perpendicular if their slopes are negative reciprocals (5/7 goes to -7/5).
Assuming in the form y = mx + b:
Solving for b:
-7/5(-5) + x = 11
(5's & negatives cancel out...)
7 + x = 11
x = 4
y = -7/5x + 4
W = L - 15
WL = 76
(L-15)L = 76
L^2 - 15L - 76 = 0
(L-19)(L+4) = 0
L = 19 or -4
Since a length cannot be negative, the length must be 19 units.
The width is 19-15=4 units.
For 9, the answer would be -4
And for 10, the answer should be 47°