Determine if the following pair of functions is linearly independent or dependent on (-infinity sign, infinity sign): f(t) = e^(3x) and g(t)=xe^(3x)
"(-infinity sign, infinity sign)"
means the interval from -∞ to +∞.
If you have f(t) and g(t) and need to know if they are linearly independent, you can calculate the Wronskian, given by the determinant:
W=
| f(t) g(t) |
|f'(t) g'(t)|
If the Wronskian is non-zero for any value of t, then the two functions are independent.
In this case, the Wronskian turns out to be e^(6t) which is non-zero for all values of t.
To determine if the pair of functions f(t) = e^(3x) and g(t) = xe^(3x) is linearly independent or dependent on (-∞, ∞), we need to check if there exist any constants a and b (not both zero) such that the linear combination a*f(t) + b*g(t) equals the zero function for all x in the given interval.
To proceed, let's calculate the linear combination:
a*f(t) + b*g(t) = a*e^(3x) + b*xe^(3x)
Now we need to determine if there exist any constants a and b (not both zero) such that a*e^(3x) + b*xe^(3x) = 0 for all x in (-∞, ∞).
To investigate this, we'll set up an equation:
a*e^(3x) + b*xe^(3x) = 0
To get a clearer picture, let's analyze the equation term by term:
Term 1: a*e^(3x)
Term 2: b*xe^(3x)
For this equation to hold for all x in the given interval, both terms must be equal to zero individually.
Considering Term 1: a*e^(3x) = 0
This equation states that a times the exponential function e^(3x) equals zero. As e^(3x) is always positive (since x ranges from -∞ to ∞), the only way this can hold is if a equals zero.
Considering Term 2: b*xe^(3x) = 0
This equation states that b times x times e^(3x) equals zero. To satisfy this equation for all x, b = 0 or x = 0.
Since a and b both equal zero are the only solutions to the equation, we can conclude that the pair of functions f(t) = e^(3x) and g(t) = xe^(3x) is linearly independent on the interval (-∞, ∞).