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Find the points of local maxima or minima for the function f(x) = x3.ex.
- a)x=-3 is a point of local maxima
- b)x=-3 is a point of local minima
- c)x=0 is a point of local maxima
- d)x=0 is a point of local minima
Correct answer is option 'B'. Can you explain this answer?
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Find the points of local maxima or minima for the function f(x)...
Solution:
The given function is f(x) = x3.ex.
To find the points of local maxima or minima, we need to find the critical points of the function.
Critical points: The points where the derivative of the function is either zero or does not exist.
f'(x) = 3x2.ex + x3.ex
Let f'(x) = 0, then
3x2.ex + x3.ex = 0
x2(ex + x) = 0
x = 0 or x = -ex
Now, we need to check the nature of critical points using the second derivative test.
f''(x) = 6x.ex + 6x2.ex + 2x3.ex
At x = 0,
f''(0) = 0
Thus, x = 0 is not a point of local maxima or minima.
At x = -ex,
f''(-ex) = 6(-ex).ex + 6(-ex)2.ex + 2(-ex)3.ex
f''(-ex) = -2ex3 < />
Thus, x = -ex is a point of local maxima.
Hence, option B is the correct answer.
Note: The second derivative test is used to determine the nature of critical points. If f''(x) > 0, then the critical point is a point of local minima. If f''(x) < 0,="" then="" the="" critical="" point="" is="" a="" point="" of="" local="" maxima.="" if="" f''(x)="0," then="" the="" test="" is="" inconclusive.="" 0,="" then="" the="" critical="" point="" is="" a="" point="" of="" local="" maxima.="" if="" f''(x)="0," then="" the="" test="" is="" />
The given function is f(x) = x3.ex.
To find the points of local maxima or minima, we need to find the critical points of the function.
Critical points: The points where the derivative of the function is either zero or does not exist.
f'(x) = 3x2.ex + x3.ex
Let f'(x) = 0, then
3x2.ex + x3.ex = 0
x2(ex + x) = 0
x = 0 or x = -ex
Now, we need to check the nature of critical points using the second derivative test.
f''(x) = 6x.ex + 6x2.ex + 2x3.ex
At x = 0,
f''(0) = 0
Thus, x = 0 is not a point of local maxima or minima.
At x = -ex,
f''(-ex) = 6(-ex).ex + 6(-ex)2.ex + 2(-ex)3.ex
f''(-ex) = -2ex3 < />
Thus, x = -ex is a point of local maxima.
Hence, option B is the correct answer.
Note: The second derivative test is used to determine the nature of critical points. If f''(x) > 0, then the critical point is a point of local minima. If f''(x) < 0,="" then="" the="" critical="" point="" is="" a="" point="" of="" local="" maxima.="" if="" f''(x)="0," then="" the="" test="" is="" inconclusive.="" 0,="" then="" the="" critical="" point="" is="" a="" point="" of="" local="" maxima.="" if="" f''(x)="0," then="" the="" test="" is="" />
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Find the points of local maxima or minima for the function f(x)...
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Find the points of local maxima or minima for the function f(x) = x3.ex.a)x=-3 is a point of local maximab)x=-3 is a point of local minimac)x=0 is a point of local maximad)x=0 is a point of local minimaCorrect answer is option 'B'. Can you explain this answer?
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