Answer: No. f(x), g(x), & h(x) are linearly Dependent.
<u>Step-by-step explanation:</u>
f(x) = eˣ - cos(x) g(x) = eˣ + cos(x) h(x) = cos(x)
Create a 3 x 3 matrix where row 1 is x = 0, row 2 is x = π/2, row 3 = π
then find the determinant.
If det = 0 ⇒ dependent.
If det ≠ 0 ⇒ independent.
![det \left[\begin{array}{ccc}f(0)&g(0)&h(0)\\f(\frac{\pi}{2})&g(\frac{\pi}{2})&h(\frac{\pi}{2})\\f(\pi)&g(\pi)&h(\pi)\end{array}\right]](https://tex.z-dn.net/?f=det%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Df%280%29%26g%280%29%26h%280%29%5C%5Cf%28%5Cfrac%7B%5Cpi%7D%7B2%7D%29%26g%28%5Cfrac%7B%5Cpi%7D%7B2%7D%29%26h%28%5Cfrac%7B%5Cpi%7D%7B2%7D%29%5C%5Cf%28%5Cpi%29%26g%28%5Cpi%29%26h%28%5Cpi%29%5Cend%7Barray%7D%5Cright%5D)
![det \left[\begin{array}{ccc}e^0-cos(0)&e^0+cos(0)&cos(0)\\e^{\frac{\pi}{2}}-cos(\frac{\pi}{2})&e^{\frac{\pi}{2}}+cos(\frac{\pi}{2})}&cos(\frac{\pi}{2})\\e^\pi-cos(\pi)&e^\pi+cos(0\pi)&cos(0\pi)\end{array}\right] =0](https://tex.z-dn.net/?f=det%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7De%5E0-cos%280%29%26e%5E0%2Bcos%280%29%26cos%280%29%5C%5Ce%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D-cos%28%5Cfrac%7B%5Cpi%7D%7B2%7D%29%26e%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%2Bcos%28%5Cfrac%7B%5Cpi%7D%7B2%7D%29%7D%26cos%28%5Cfrac%7B%5Cpi%7D%7B2%7D%29%5C%5Ce%5E%5Cpi-cos%28%5Cpi%29%26e%5E%5Cpi%2Bcos%280%5Cpi%29%26cos%280%5Cpi%29%5Cend%7Barray%7D%5Cright%5D%20%3D0)
Since determinant = 0, they are linearly dependent.
There are an infinite number of possibilities to make them independent. <em>Any change that makes the determinant ≠ 0</em>
One example would be to make h(x) = sin(x)