Final answer:
To determine if the given linear transformation t is invertible, we need to check if it is one-to-one and onto. For a linear transformation to be invertible, it must be one-to-one and onto. After checking both conditions, we find that t is not one-to-one and not onto, therefore it is not invertible. So the correct answer is B) t is not invertable.
Step-by-step explanation:
The linear transformation t is given by t(f(x)) = f′′(x) 2f′(x) − f(x).
To determine if t is invertible, we need to check if it is one-to-one and onto. For a linear transformation to be invertible, it must be one-to-one and onto.
For t to be one-to-one, the equation t(f(x)) = 0 should only have the trivial solution, which means that if t(f(x)) = t(g(x)), then f(x) = g(x) for all x in the domain.
For t to be onto, given any vector y in the codomain p2(r), there must exist a vector x in the domain p2(r) such that t(x) = y. In other words, t must be able to map every element of the codomain p2(r) to some element in the domain p2(r).
Let's check if t is one-to-one and onto.
One-to-one:
To check if t is one-to-one, we need to show that if t(f(x)) = t(g(x)), then f(x) = g(x) for all x in the domain.
Let's assume t(f(x)) = t(g(x)), then f′′(x) 2f′(x) − f(x) = g′′(x) 2g′(x) − g(x).
If we differentiate both sides of the equation, we get f′(x)² − f(x) = g′(x)² − g(x).
This implies that f′(x)² − f(x) − g′(x)² + g(x) = 0.
If we let h(x) = f′(x)² − f(x) − g′(x)² + g(x), we can see that h(x) is a quadratic polynomial.
To show that h(x) = 0 for all x in the domain, we need to show that h(x) has no roots other than x = 0. This can be done by either factoring h(x) or showing that the discriminant of h(x) is negative.
Let's try factoring h(x).
Notice that h(x) can be factored as h(x) = (f′(x) − g′(x))(f′(x) + g′(x)) − (f(x) − g(x)).
Since (f′(x) − g′(x)) and (f′(x) + g′(x)) are both polynomials of degree 2 and f(x) − g(x) is a polynomial of degree 0, the product (f′(x) − g′(x))(f′(x) + g′(x)) − (f(x) − g(x)) is a polynomial of degree 2.
Therefore, h(x) must have roots other than x = 0. Hence, t is not one-to-one.
Onto:
To check if t is onto, we need to show that for any vector y in the codomain p2(r), there exists a vector x in the domain p2(r) such that t(x) = y.
Since t maps the input function f(x) to a polynomial of degree 2, we can see that there is no way to choose a function f(x) such that t(f(x)) is a polynomial of degree 1 or 3.
Therefore, there exists vectors y in the codomain such that there is no vector x in the domain that maps to y. Hence, t is not onto.
Since t is neither one-to-one nor onto, it is not invertible. Therefore, the answer is B) t is not invertable.