Answer :
Answer:
Functions are linearly dependent (are not linearly independent.)
Step-by-step explanation:
Remember that two functions f(x), g(x) and h(x) are said linearly independent on an interval I if the only solution to the equation
[tex]\alpha f(x)+\beta g(x)+\omega h(x)=0, \ \text{for all } x\in I[/tex]
is the trivial one: α = 0, β = 0, ω = 0. If they are not linearly independent, they are called linearly dependent.
Now, let f(x), g(x) and h(x) be the functions:
[tex]f(x)=e^{3x}+\cos(5x),[/tex]
[tex]g(x)=e^{3x}-\cos(5x),[/tex]
[tex]h(x)=\cos(5x).[/tex]
Then, letting α = 1, β= -1 and ω = -2, we see that:
[tex]\alpha f(x)+\beta g(x)+ \omega h(x)=e^{3x}+\cos(5x)-e^{3x}+\cos(5x)+2\cos(5x)=0.[/tex]
Hence, the functions f(x), g(x) and h(x) are not linearly independent, or equivalently, are linearly dependent.
The system is linearly dependent as not all coefficients are zero.
How to find if three given functions are linearly independent
Let be the following linear combination:
[tex]\alpha\cdot f(x) + \beta \cdot g(x) + \gamma \cdot h(x) = 0[/tex] (1)
Where [tex]\alpha[/tex], [tex]\beta[/tex] and [tex]\gamma[/tex] are real coefficients. All functions are linearly independent when [tex]\alpha = \beta = \gamma = 0[/tex].
Then, we have the following model:
[tex]\alpha \cdot (e^{3\cdot x}+\cos 5x)+\beta \cdot (e^{3\cdot x}-\cos 5x)+\gamma \cdot \cos 5x = 0[/tex]
[tex](\alpha - \beta)\cdot e^{3\cdot x} + (\alpha - \beta +\gamma)\cdot \cos 5x = 0[/tex] (2)
Then, we have the following system of equations:
[tex]\alpha - \beta = 0[/tex] (3)
[tex]\alpha - \beta - \gamma = 0[/tex] (4)
The solution of this system is: [tex]\alpha = \beta = k[/tex], [tex]k\in \mathbb{R}[/tex], [tex]\gamma = 0[/tex]. Hence, we conclude that the system is linearly dependent as not all coefficients are zero. [tex]\blacksquare[/tex]
To learn more on linear independence, we kindly invite to check this verified question: https://brainly.com/question/16967734