Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:
(i) `f(x)=x^2`
(ii) `g(x)=x^3-3x`
(iii) ` h(x)=sinx+cosx ,0ltxltpi/ 2`
(iv)`f(x) = sin x - cos x `,` 0 lt x lt 2pi `
(v) f(x) = `x^3-6x^2+9x+15`
(vi) g(x) =` x/2 +2/x` , `x gt 0`
(vii) g(x) = `1/(x^2 +2) `
(viii) `f(x) = x sqrt(1-x)` , `x gt 0`

Explanation

(i) `f(x)=x^2`
`f^'(x)=2x=0 =>x=0` is critical point of function
At `x=0` , Minimum value of function=0

(ii) `g(x)=x^3-3x`
`g^'(x)=3x^2-3=0=>3x^2=3=>x^2=1=>x=+-1` is the critical point of function
Now, `g^('')(x)=6x`
`g^('')(1)=6 gt 0` , `x=1` is local minima point
`g^('')(-1)=-6 lt 0` , `x=-1` is local maxima point
Maximum value at `x=-1` should be `2`
Minimum value at `x=1` should be `-2`
(iii) `h(x)=sinx+cosx`
`h^'(x)=cosx-sinx=0=> tanx=1`
maximum=`1/sqrt(2)+1/sqrt(2)=sqrt(2)`
minimum=`-1/sqrt(2)-1/sqrt(2)=-sqrt(2)`
But in this case , `0lt x ltpi/2`
minimum at x=0 should be 1.
(iv) `f(x)=sinx-cosx`
`f^'(x)=cosx+sinx=0=>tanx=-1`
`x=(3pi)/4,(7pi)/4`
Getting, `f^('')(x)=cosx-sinx`
`f^('')((3pi)/4)=cos((3pi)/4)-sin((3pi)/4)=-1/sqrt(2)-1/sqrt(2)=-sqrt(2)`
i.e., `f^('')((3pi)/4)` is negative, so at `x=(3pi)/4,f(x)` is local maxima
Hence, local maximum value =`1/sqrt(2)+1/sqrt(2)=sqrt(2)`
`f^('')((7pi)/4)=cos((7pi)/4)-sin((7pi)/4)=1/sqrt(2)+1/sqrt(2)`
`f^('')((7pi)/2)` is positive so `x=(7pi)/4` is local minima
Hence, local minimum value =`-1/sqrt(2)-1/sqrt(2)=-sqrt(2)`
(v) `f(x)=x^3-6x^2+9x+15`
`f^'(x)=3x^2-12x+9=0=> 3(x^2-4x+3)=0`
`=3(x-1)(x-3)=0`
`x=1,3`
Now, `f^('')(x)=6x-12=6(x-2)`
`f^('')(1)=6(1-2)=-6lt0`
`f^('')(3)=6(3-2)=6>0`
Therefore , by second derivative test , `x=1` is a point of local maxima and the local maximum value of f at `x=1` at `F(1)=1-6+9+15=19.`
However, x=3 is a point of local minima and the local minimum value of f at `x=3` is `f(3)=27-54+27+15=15`
(vi) `g(x)=x/2+2/x,xgt0`
`g^'(x)=1/2-2/x^2`
Now, `g^'(x)=0` gives `2/x^2=1/2=>x^2=4=>x=+-2`
Since `x gt 0`, we take `x=2`
Now, `g^('')(x)=4/x^3,g^('')(2)=4/2^3=1/2gt0`
Therefore, by second derivative test, `x=2` is a point of local minima and the local minimum value of g at `x=2` is `g(2)=2/2+2/2=2`
(vii) `g(x)=1/(x^2+2)`
`g^'(x)=-(2x)/(x^2+2)^2`
`g^'(x)=0=> -(2x)/(x^2+2)^2=0=>x=0`
Now for values close to `x=0,g^'(x) gt0` and to the left of 0 , Also, for values close to x=0 and to the right of 0, `g^'(x)lt0`
Therefore, by first derivative test, `x=0` is a point of local maxima and the local maximum value of `g(0)` is `1/(0+2)=1/2`