4\n(i) the diagram below shows the graph of ( y = f(x) ) for ( x leq 0 ). it is given that ( f(x) ) is an…

4\n(i) the diagram below shows the graph of ( y = f(x) ) for ( x leq 0 ). it is given that ( f(x) ) is an odd\nfunction such that ( f(-x)=-f(x) ) and the graph cuts the ( x )-axis at ( x = -7, x = -2.5 ) and\n( x = 0 ).\n(a) sketch the graph of ( y = f(x) ) for ( x>0 ). label the axial intercepts clearly. 1\n(b) given further that ( int_{-2.5}^{-7} f(x) d x=-p ), and the area bounded by ( y = f(x) ) and ( x )-axis\nfrom ( x=-7 ) to ( x = 0 ) is ( q ) units ( ^{2} ), where ( p ) and ( q ) are positive constants. find\n( int_{0}^{2.5} k f(-x) d x ) in terms of ( k, p ) and/or ( q ). 2

4\n(i) the diagram below shows the graph of ( y = f(x) ) for ( x leq 0 ). it is given that ( f(x) ) is an odd\nfunction such that ( f(-x)=-f(x) ) and the graph cuts the ( x )-axis at ( x = -7, x = -2.5 ) and\n( x = 0 ).\n(a) sketch the graph of ( y = f(x) ) for ( x>0 ). label the axial intercepts clearly. 1\n(b) given further that ( int_{-2.5}^{-7} f(x) d x=-p ), and the area bounded by ( y = f(x) ) and ( x )-axis\nfrom ( x=-7 ) to ( x = 0 ) is ( q ) units ( ^{2} ), where ( p ) and ( q ) are positive constants. find\n( int_{0}^{2.5} k f(-x) d x ) in terms of ( k, p ) and/or ( q ). 2

Answer

Part (a)

Brief Explanations:

Since (y = f(x)) is an odd function, its graph is symmetric about the origin. For (x>0), if ((x,y)) is on the graph of (y = f(x)) for (x < 0), then ((-x,-y)) is on the graph for (x>0). The (x -)intercepts for (x>0) are (x = 2.5) and (x=7) (because if (f(a)=0) for (a<0), then (f(-a)=-f(a) = 0) for (a>0)).

Answer:

Sketch the graph of (y = f(x)) for (x>0) with (x -)intercepts at (x = 2.5) and (x = 7). The graph for (x>0) is the reflection of the graph for (x<0) about the origin.

Part (b)

Explanation:

Step1: Use the property of odd - function and substitution

Let (u=-x), then (du=-dx). When (x = 0), (u = 0); when (x = 2.5), (u=-2.5). So (\int_{0}^{2.5}kf(-x)dx=-k\int_{0}^{-2.5}f(u)du=k\int_{-2.5}^{0}f(u)du).

Step2: Use the relationship between definite - integrals and areas

We know that (\int_{-7}^{-2.5}f(x)dx=-p) and (\int_{-7}^{0}f(x)dx = q). By the property of definite - integrals (\int_{-7}^{0}f(x)dx=\int_{-7}^{-2.5}f(x)dx+\int_{-2.5}^{0}f(x)dx). Then (\int_{-2.5}^{0}f(x)dx=q + p).

Answer:

(\int_{0}^{2.5}kf(-x)dx=k(q + p))