Question

Determine whether the pairs of functions: f_(1)(x)=e^(3x) and f_(2)(x)=e^(-3x) are linearly dependent or linearly independent.

Answer 1

the only way for a*f₁(x) + b*f₂(x) to equal zero for all values of x is if a = b = 0. This means that the pair of functions f₁(x) = e^(3x) and f₂(x) = e^(-3x) are linearly independent.

Step-by-step explanation:

The pair of functions f₁(x) = e^(3x) and f₂(x) = e^(-3x) are linearly independent. To determine this, we need to check if there exist constants a and b, not both zero, such that a*f₁(x) + b*f₂(x) = 0 for all values of x.

Let's assume a and b are constants and substitute f₁(x) and f₂(x) into the equation:

a*e^(3x) + b*e^(-3x) = 0

To find the values of a and b, we can try substituting different values of x. Let's start with x = 0:

a*e^(0) + b*e^(0) = a + b = 0

From this equation, we can see that a = -b. Therefore, if a and b are not both zero, then a = -b = k, where k is a non-zero constant.

Now, substituting this value of a into the equation, we have:

k*e^(3x) - k*e^(-3x) = k*(e^(3x) - e^(-3x))

We know that e^(3x) and e^(-3x) are always positive. Therefore, for the expression k*(e^(3x) - e^(-3x)) to equal zero for all values of x, k must be zero. But this contradicts our assumption that k is non-zero.

Answer 2

After calculating the Wronskian determinant for functions f_(1)(x)=e^(3x) and f_(2)(x)=e^(-3x), we found it to be 0. Hence, the functions are linearly dependent.

Step-by-step explanation:

Determining Linear Dependence or Independence

To determine whether the pairs of functions f_(1)(x)=e^(3x) and f_(2)(x)=e^(-3x) are linearly dependent or independent, we use the Wronskian which is a determinant used to test for linear dependence of two functions. For two functions to be linearly dependent, there must exist constants C1 and C2, not both zero, such that C1*f_(1)(x) + C2*f_(2)(x) = 0 for all x in the domain of the functions.

To apply the test, we calculate the Wronskian given by:

W(f_(1), f_(2)) = |

f_(1)(x)f_(2)(x)
f'_(1)(x)f'_(2)(x)


| = |

e^(3x)e^(-3x)
3e^(3x)-3e^(-3x)

|

Computing the determinant, we find:

W(f_(1), f_(2)) = (e^(3x) * -3e^(-3x)) - (3e^(3x) * e^(-3x)) = -3 + 3 = 0

However, for functions f_(1) and f_(2) to be linearly dependent, the Wronskian must be zero at all points x. Since our Wronskian is zero for all x, f_(1)(x) and f_(2)(x) are indeed linearly dependent.