Question

Consider the following functions. f1(x) = 0, f2(x) = x, f3(x) = ex g(x) = c1f1(x) + c2f2(x) + c3f3(x) Solve for c1, c2, and c3 so that g(x) = 0 on the interval (−[infinity], [infinity]). If a nontrivial solution exists, state it. (If only the trivial solution exists, enter the trivial solution {0, 0, 0}.) {c1, c2, c3} = Incorrect: Your answer is incorrect. Determine whether f1, f2, f3 are linearly independent on the interval (−[infinity], [infinity]). linearly dependent linearly independent Correct: Your answer is correct.

Answer 1

Answer:

$(c_1,c_2,c_3)=(0,0,0)$

Linearly independent

Step-by-step explanation:

The given functions are:

$f_1(x)=0,f_2(x)=x,f_3(x)=e^x$

$g(x)=c_1f_1(x)+c_2f_2(x)+c_3f_3(x)$

We want to solve for $c_1,c_2,$ and $c_3$  so that $g(x)=0$ on the interval $(-\infty,\infty)$.

Let us substitute to get:

$c_1\cdot 0+c_2\cdot x+c_3\cdot e^x=0$

This implies that:

$0+c_2x+c_3e^x=0\\\implies c_2x+c_3e^x=0$--->eqn1

Let us differentiate each ter wrt x to get:

$c_2+c_3e^x=0$---->eqn2

Make $c_2$ the subject in equation (2) to get:

$c_2=-c_3e^x$---->eqn3

Substitute $c_2=-c_3e^x$ into equation (1) to get:

$-c_3xe^x+c_3e^x=0$

Factor nicely to get:

$c_3(1-x)e^x=0$

Note that we are after constants that will this equation true for all x. The only way this could be possible is when $c_3=0$

Put $c_3=0$ into eqn3 to get:

$c_2=-\cdot 0\cdot e^x\\\implies c_2=0$

Let us substitute everything into g(x) to find $c_1$

$c_1\cdot 0+0\cdot x+0\cdot e^x=0$

This implies that:

$c_1\cdot 0+0+0=0\\\implies c_1\cdot 0=0$

By the zero product principle , for r $c_1\cdot 0=0$, we must have either $o=o$ or $c_1=0$.

Since $0=0$ is true, $c_1=0$ is also true

Since $c_1f_1(x)+c_2f_2(x)+c_3f_3(x)=0$  for all x, if $c_1=0,c_2=0,c_3=0$, the functions $f_1(x)=0,f_2(x)=x,f_3(x)=e^x$ are linearly independent.