For the function, find the points on the graph at which the tangent line is horizontal. If none exist, state that fact. SHOW W
ORK PLEASE!
2 answers:
9514 1404 393
Answer:
A. (-√2, 3+(4√2)/3), (√2, 3-(4√2)/3)
Step-by-step explanation:
The tangents are horizontal where the derivative is zero. The derivative of the given function is ...
y' = x² -2
This is zero when ...
0 = x² -2
2 = x²
±√2 = x . . . . x-values where the derivative is zero
__
The corresponding y-values are ...
y = (1/3x² -2)x +3
y = (1/3(2) -2)(±√2) +3 = 3 ∓ (4√2)/3
The turning points are (-√2, 3+(4√2)/3) and (√2, 3-(4√2)/3).
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