Find the points of local maxima or local minima, if any, of the following function, using the first derivative test. Also, find the local maximum or local minimum values, as the case may be: `f(x)=x/2+2/x ,x >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`