Evaluate $T$* at the given vector in $ P_1(R)$

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For problem 3(c)
enter image description here



I get that $T(at + b) = (at +b)' + 3(at + b) = a + 3at + 3b$. Then I get that



$langle a + 3at + 3b, 4 - 2t rangle$.



By definition, $langle f,g rangle = int_-1^1 -6at^2 + (10a -6b)t + 4a +12b ,dt$,



but the answer I find online gives me



$int_-1^1 -6at^2 + (-2a + 6b)t + 4a +12b ,dt$.



I was wondering where I went wrong?




linear-algebra




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  • Perhaps you didn’t. It sure looks like the online answer is missing a term.
    – amd
    Jul 20 at 23:24










  • thanks, I wasn't sure if I was missing something important.
    – K.M
    Jul 20 at 23:27














up vote
0
down vote

favorite












For problem 3(c)
enter image description here



I get that $T(at + b) = (at +b)' + 3(at + b) = a + 3at + 3b$. Then I get that



$langle a + 3at + 3b, 4 - 2t rangle$.



By definition, $langle f,g rangle = int_-1^1 -6at^2 + (10a -6b)t + 4a +12b ,dt$,



but the answer I find online gives me



$int_-1^1 -6at^2 + (-2a + 6b)t + 4a +12b ,dt$.



I was wondering where I went wrong?




linear-algebra




share|cite|improve this question



















  • Perhaps you didn’t. It sure looks like the online answer is missing a term.
    – amd
    Jul 20 at 23:24










  • thanks, I wasn't sure if I was missing something important.
    – K.M
    Jul 20 at 23:27












up vote
0
down vote

favorite









up vote
0
down vote

favorite











For problem 3(c)
enter image description here



I get that $T(at + b) = (at +b)' + 3(at + b) = a + 3at + 3b$. Then I get that



$langle a + 3at + 3b, 4 - 2t rangle$.



By definition, $langle f,g rangle = int_-1^1 -6at^2 + (10a -6b)t + 4a +12b ,dt$,



but the answer I find online gives me



$int_-1^1 -6at^2 + (-2a + 6b)t + 4a +12b ,dt$.



I was wondering where I went wrong?




linear-algebra




share|cite|improve this question











For problem 3(c)
enter image description here



I get that $T(at + b) = (at +b)' + 3(at + b) = a + 3at + 3b$. Then I get that



$langle a + 3at + 3b, 4 - 2t rangle$.



By definition, $langle f,g rangle = int_-1^1 -6at^2 + (10a -6b)t + 4a +12b ,dt$,



but the answer I find online gives me



$int_-1^1 -6at^2 + (-2a + 6b)t + 4a +12b ,dt$.



I was wondering where I went wrong?





linear-algebra





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 20 at 23:15









K.M

480312




480312











  • Perhaps you didn’t. It sure looks like the online answer is missing a term.
    – amd
    Jul 20 at 23:24










  • thanks, I wasn't sure if I was missing something important.
    – K.M
    Jul 20 at 23:27
















  • Perhaps you didn’t. It sure looks like the online answer is missing a term.
    – amd
    Jul 20 at 23:24










  • thanks, I wasn't sure if I was missing something important.
    – K.M
    Jul 20 at 23:27















Perhaps you didn’t. It sure looks like the online answer is missing a term.
– amd
Jul 20 at 23:24




Perhaps you didn’t. It sure looks like the online answer is missing a term.
– amd
Jul 20 at 23:24












thanks, I wasn't sure if I was missing something important.
– K.M
Jul 20 at 23:27




thanks, I wasn't sure if I was missing something important.
– K.M
Jul 20 at 23:27















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