Let f(x)= x - 3 + (2/x+1) determine the intervals where f(x) is increasing, inte

Question

Let f(x)= x - 3 + (2/x+1) determine the intervals where f(x) is increasing, intervals where f(x) is decreasing, intervals where f(x) is concave up, concave down, all points whicj give a local maximum and minumum value, and all points of inflection.

Explanation / Answer

For given f(x) = x - 3 + (2/x+1) on differentiating f'(x) = 1- 2 / x^2 = 0 ==> x = + sqrt ( 2) , - sqrt ( 2 ) for f(x) to be increasing , f'(x) > 0 for f(x) to be increasing x belongs to ( - infinity , -sqrt (2) ) U ( sqrt ( 2) , infinity ) for f(x) to be decreasing x belongs to ( - sqrt ( 2) , sqrt ( 2 ) ) here , f''(x) = 4/ x^3 since f''(x) > 0 for x belongs to ( 0 , infinity ) ---> f(x) concave up f''(x) < 0 for x belongs to ( -infinity , 0 ) ---> f(x) concave downwards since @ x= 0 f''(x) is changing it's sign therefore , x= 0 is a point of inflexion but since it is not in domain of f(x) therefore no point of inflexion . @ x = sqrt ( 2) , f''(x) > 0 --> x = sqrt ( 2) { local minima ) @ x = - sqrt ( 2) , f''(x) < 0 --> x = -sqrt ( 2) { local maxima )

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