Question

The function f is defined by

 for .

    i.      Express in the form of where a and b are constants.

  ii.     State the range of . 

The function g is defined by

 for .

 iii.       State the largest value of k for which g has an inverse. 

 iv.     Given that g has an inverse, find an expression for .

Solution

i.
 

We have the function;

We have the expression;

We use method of “completing square” to obtain the desired form. We complete the square for the  terms which involve .

We have the algebraic formula;

For the given case we can compare the given terms with the formula as below;

Therefore, we can deduce that;

Hence, we can write;

To complete the square, we can add and subtract the deduced value of ;

Hence;

   ii.
 

Standard form of quadratic equation is;

The graph of quadratic equation is a parabola. If (‘a’ is positive) then parabola opens upwards  and its vertex is the minimum point on the graph. If (‘a’ is negative) then parabola  opens downwards and its vertex is the maximum point on the graph.

We recognize that given curve , is a parabola opening upwards.

Vertex form of a quadratic equation is;

The given curve , as demonstrated above can be written in vertex form as;

Coordinates of the vertex are .Since this is a parabola opening upwards the vertex is the  minimum point on the graph. Here y-coordinate of vertex represents least value of and x- coordinate of vertex represents corresponding value of . 

For the given case, vertex is . Therefore, least value of  is and corresponding value of   is 3.

Hence range of ;

 iii.
 

The function is defined by for , where k is a constant.

We are required to state the largest value of k for which g has an inverse.

For a given function the inverse function exists only if is one-one function.

A one-one function has only one value of against one value of . A function is one-one function if  it passes horizontal line test i.e., horizontal line passes only through one point of function. However,  if a function is not one-one, we can make it so by restricting its domain.

We have;

 for

The given function is a parabola and parabola does not pass horizontal line test, so it is not a one- one function. However, we can make it so by restricting its domain.

Standard form of quadratic equation is;

The graph of quadratic equation is a parabola. If (‘a’ is positive) then parabola opens upwards  and its vertex is the minimum point on the graph. If (‘a’ is negative) then parabola  opens downwards and its vertex is the maximum point on the graph.

We recognize that given curve   , is a parabola opening upwards. 

Vertex form of a quadratic equation is;

The given curve , as demonstrated in (i) can be written in vertex form as; 

Coordinates of the vertex are .Since this is a parabola opening upwards the vertex is the  minimum point on the graph. Here y-coordinate of vertex represents least value of and x- coordinate of vertex represents corresponding value of . 

For the given case, vertex is . Therefore, least value of  is and corresponding value of   is .

The given function is a parabola and parabola does not pass horizontal line test, so it is not a one- one function. However, we can make it so by restricting its domain around its line of symmetry.

The line of symmetry of a parabola is a vertical line passing through its vertex. 

Vertex of the given parabola, as demonstrated above is . Hence the vertical line passing  through the vertex is . By restricting the domain of the given function to , we can make it  one-one function. On both sides of the vertical line given function is a one-one function.

Therefore, for , given function is a one-one function. Hence;

 iv.
 

To find the inverse of a given function we need to write it in terms of rather than in terms of . 

As demonstrated in (i), we can write the given function as;

Interchanging ‘x’ with ‘y’;

As demonstrated in (iii), given function is a one-one function, has inverse, for; 

Therefore, only possible option is;