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Mandarinka [93]
2 years ago
4

Let f ( x ) = e x − cos x , f(x)=ex−cos⁡x, g ( x ) = e x + cos x , g(x)=ex+cos⁡x, and h ( x ) = cos x . h(x)=cos⁡x. Are f ( x )

, f(x), g ( x ) , g(x), and h ( x ) h(x) linearly independent? If so, show it, if not, find a linear combination that works.
Mathematics
1 answer:
MArishka [77]2 years ago
4 0

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]

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

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)

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