Input each of the following functions in Maple (f(x) = x(x – 2)

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Input each of the following functions in Maple (f(x) = x(x – 2), g(x) = x – 3, h(x)= x -1 ) and then compute and simplify each of the following: a) figh(x))) b) ho f(x))) c) fhg x))) Question 5 Input f(x) = (1 + x’)’into Maple. Then create a plot (over -3 s Xs 3) containing the graphs of: a) f(x), f(x-1), and f(x+1) b) f(x), f(2x), and f(8x) Question 6 l, if x < 0=”” f(x)=”1-X,” if=”” 0=”” sxs=”” 2=”” (x+1,=”” if=”” x=””> 2 Using Maple, plot

The Correct Answer and Explanation is :

To address the given tasks in Maple, here’s a structured approach for each function and computation.

Task 1: Function Definitions and Simplification

Given the functions:

  • ( f(x) = x(x – 2) )
  • ( g(x) = x – 3 )
  • ( h(x) = x – 1 )

The following operations need to be performed:

a) ( f \circ g \circ h(x) )

This means composing ( f(x) ), ( g(x) ), and ( h(x) ), so we first apply ( h(x) ), then ( g(x) ), and finally ( f(x) ).

  1. Start with ( h(x) = x – 1 ).
  2. Apply ( g ) to ( h(x) ): ( g(h(x)) = (x – 1) – 3 = x – 4 ).
  3. Finally, apply ( f ) to ( g(h(x)) ): ( f(x – 4) = (x – 4)(x – 4 – 2) = (x – 4)(x – 6) ).

So, ( f \circ g \circ h(x) = (x – 4)(x – 6) ).

b) ( h \circ f(x) )

This means applying ( f(x) ) first and then applying ( h ) to the result.

  1. Start with ( f(x) = x(x – 2) = x^2 – 2x ).
  2. Apply ( h ) to ( f(x) ): ( h(f(x)) = (x^2 – 2x) – 1 = x^2 – 2x – 1 ).

So, ( h \circ f(x) = x^2 – 2x – 1 ).

c) ( f \circ h \circ g(x) )

This means applying ( g(x) ) first, then ( h(x) ), and finally ( f(x) ).

  1. Start with ( g(x) = x – 3 ).
  2. Apply ( h ) to ( g(x) ): ( h(g(x)) = (x – 3) – 1 = x – 4 ).
  3. Finally, apply ( f ) to ( h(g(x)) ): ( f(x – 4) = (x – 4)(x – 6) ).

So, ( f \circ h \circ g(x) = (x – 4)(x – 6) ).

Task 2: Plotting Functions

Now, you are asked to plot the function ( f(x) = (1 + x^2)’ ) and generate two different sets of plots.

a) Plot ( f(x) ), ( f(x – 1) ), and ( f(x + 1) )

  1. Define the function: ( f(x) = (1 + x^2)’ ) (Maple will differentiate the expression).
  2. Plot ( f(x) ), ( f(x – 1) ), and ( f(x + 1) ) over the interval ([-3, 3]).

In Maple, you can use the following commands:

f := (1 + x^2):  # Define the function
df := diff(f, x);  # Differentiate the function

# Plot the functions
plots[plot]([df, subs(x = x - 1, df), subs(x = x + 1, df)], x = -3..3);

b) Plot ( f(x) ), ( f(2x) ), and ( f(8x) )

  1. Again, define ( f(x) = (1 + x^2)’ ).
  2. Create a plot with ( f(x) ), ( f(2x) ), and ( f(8x) ) over the interval ([-3, 3]).

In Maple, you can use the following commands:

# Plot the functions
plots[plot]([df, subs(x = 2*x, df), subs(x = 8*x, df)], x = -3..3);

Task 3: Piecewise Function Plot

You are given a piecewise function:

  • ( f(x) = 1 – x ) for ( x < 0 )
  • ( f(x) = x + 1 ) for ( 0 \leq x \leq 2 )
  • ( f(x) = x ) for ( x > 2 )

To plot this function, use Maple’s piecewise function and create a graph over an appropriate domain.

f := piecewise(x < 0, 1 - x, 0 <= x and x <= 2, x + 1, x > 2, x);

# Plot the piecewise function
plots[plot](f, x = -3..3);

Explanation:

In this task, we plotted the given piecewise function over the interval ( -3 \leq x \leq 3 ). The graph will show three different behaviors:

  1. A line ( f(x) = 1 – x ) when ( x < 0 ),
  2. A line ( f(x) = x + 1 ) when ( 0 \leq x \leq 2 ),
  3. A line ( f(x) = x ) when ( x > 2 ).

The plot will clearly demonstrate how the function behaves differently depending on the value of ( x ), with discontinuities at ( x = 0 ) and ( x = 2 ). The graph provides a visual representation of the piecewise nature of the function.

Conclusion:

Using Maple, we can efficiently compute function compositions, differentiate functions, and plot complex piecewise functions. The piecewise function will help to understand how different expressions apply to different regions of ( x ), making it a valuable tool for modeling real-world phenomena with different behaviors in different intervals.

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