Question -
Answer -
(i) f(x)= x2
Thus, x =0 is the only critical point which could possibly be the point of local maximaor local minima of f.
We have
, which is positive.
Therefore, by second derivative test, x =0 is a point of local minima and local minimum value of f at x =0 is f(0) = 0.
(ii) g(x)= x3 −3x
By second derivative test, x = 1 is a point of local minima and local minimum valueof g at x = 1 is g(1)= 13 − 3 = 1 − 3 = −2. However,
x = −1 is a point of local maxima and local maximumvalue of g at
x = −1 is g(1) = (−1)3 −3 (− 1) = − 1 + 3 = 2.
(iii) h(x) =sinx +cosx, 0< x <
Therefore, by secondderivative test,
is a point of localmaxima and the local maximum value of h at is 
Therefore, by secondderivative test,
is a point of localmaxima and the local maximum value of f at
is
However,
is a point of localminima and the local minimum value of f atis 
(v) f(x)= x3 −6x2 +9x + 15
Therefore, by second derivative test, x =1 is a point of local maxima and the local maximum value of f at x = 1 is f(1) = 1 − 6 + 9 + 15 =19. However, x = 3 is a point of local minima and the local minimumvalue of f at x = 3 is f(3) = 27 − 54 + 27 + 15 = 15.
(vi) 
Since x >0, we take x = 2.
Therefore, by secondderivative test, x = 2 is a point of local minima and the localminimum value of g at x =2 is g(2) =
(vii) 
Now, for values closeto x = 0 and to the left of 0,
Also, for values closeto x = 0 and to the right of 0,
Therefore, by firstderivative test, x = 0 is a point of local maxima and the localmaximum value of
(viii) 
Therefore, by secondderivative test,
is a point of local maximaand the local maximum value of f at is