Question

2. Find the maximum and minimum values, if any,of the following functionsgiven by(i) f(x)- lx +21-1(ili) h(x)-sin (2x)5v)h(x) = χ + 1, XE(-1,1(ii) gx)3(iv) f(x) Isin 4x +31

Answer 1

(i)

Given function is f( x )=| x+2 |−1.

The given function can be rewritten as, f( x )={ ( x+2 )−1      ,x≥−2 −( x+2 )−1   ,x<−2

Consider f( x )=−( x+2 )−1, when x<−2,

The derivative of this function will be.

f ′ ( x )=−1

Hence for any x<−2, f'( x ) is negative.

Consider f( x )=( x+2 )−1 when x≥−2,

f ′ ( x )=1

Hence for any x≥−2, f'( x ) is positive.

Since f'( x )changes from negative to positive at x=−2, thus x=−2 is the point of minima.

Minimum value of the function is,

f( −2 )=| −2+2 |−1 =−1

Therefore, minimum value of function is −1 and maximum value does not exist as the value of the function goes on increasing by increasing the value of x.

(ii)

Given function is g( x )=−| x+1 |+3.

The given function can be rewritten as, g( x )={ −( x+1 )+3      ,x≥−1 ( x+1 )+3        ,x<−1

Consider g( x )=( x+1 )+3, for, x<−1

The derivative of this function will be,

g'( x )=1

Consider g( x )=−( x+1 )+3, for x≥−1,

The derivative of this function is,

g'( x )=−1

Since the derivative of the function changes its sign from positive to negative at x=−1

Thus, x=−1 is the point of maxima. There is no point of minima as the value of the function goes on decreasing for higher value of x.

Maximum value of function is,

| −1+1 |+3=3

(iii)

Given function is h( x )=sin( 2x )+5.

As −1≤sinθ≤1.

So,

−1≤sin2x≤1 −1+5≤sin2x+5≤1+5 4≤sin2x+5≤6

Thus maximum value of the given function is 6and the minimum value is 4.

(iv)

Given function is f( x )=| sin4x+3 |.

As −1≤sinθ≤1.

So,

−1≤sin4x≤1 −1+3≤sin4x+3≤1+3 2≤sin4x+3≤4 | 2 |≤| sin4x+3 |≤| 4 |

Thus maximum value of the given function is 4and the minimum value is 2.

(v)

Given function is h( x )=x+1, x∈( −1,1 ).

The given function h( x ) has maximum value closest to 1, and minimum value closest to −1 but it is not possible to draw that point closest to 1 and −1.

Thus, the given function has neither maximum nor minimum values.