Which statement is true about the discontinuities of the function f(x)?
1 answer:
Answer:
(a) There are asymptotes at x=3/2 and x=-1/3
Step-by-step explanation:
The denominator zeros can be found by factoring:
f(x) = (x +1)/((2x -3)(3x +1))
Neither of the denominator factors is cancelled by the numerator factor, so each represents a vertical asyptote, not a function hole.
The asymptotes are at the values of x where the denominator is zero:
2x -3 = 0 ⇒ x = 3/2
3x +1 = 0 ⇒ x = -1/3
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.
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Since, -1 = 2 is not possible, therefore there exist no such point <span>on the paraboloid such that the tangent plane is parallel to the plane 3x + 2y + 7z = 2</span>.
Given <span>the paraboloid
We can rewrite it as a scalar value function f as follows:
The normal to the </span><span>paraboloid at any point is given by:
Also, the normal to the given plane
Equating the two normal vectors, we have:
</span>
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