On considère la fonction vectoriellef:R⟶R2x⟼(2x,x2).\begin{aligned} f: \mathbb{R} & \longrightarrow \mathbb{R}^2 \\ x & \longmapsto\left(2 x, x^2\right).\end{aligned}f:Rx⟶R2⟼(2x,x2).Calculer :f(2)etf(4).f(2)\quad\text{et}\quad f(4).f(2)etf(4). I solved itMark it doneAdd to my listKeep it in your listI like this problem0 likesSolutions1Reveal solutionsAre you sure? Give it a try first.Solution by darktoasterDiscussions0 useful votesf(2)=(2⋅2,22)=(4,4)f(2) = (2 \cdot 2, 2^{2}) = (4,4)f(2)=(2⋅2,22)=(4,4)f(4)=(2⋅4,42)=(8,16)f(4) = (2 \cdot 4, 4^{2}) = (8,16)f(4)=(2⋅4,42)=(8,16) ReportFor an unclear, ambiguous, or possibly incorrect statement, please use the Discussion tab on the right. Report content that needs moderator intervention, such as dangerous, clearly non-mathematical, or plagiarized content.Submit
Solution by darktoasterDiscussions0 useful votesf(2)=(2⋅2,22)=(4,4)f(2) = (2 \cdot 2, 2^{2}) = (4,4)f(2)=(2⋅2,22)=(4,4)f(4)=(2⋅4,42)=(8,16)f(4) = (2 \cdot 4, 4^{2}) = (8,16)f(4)=(2⋅4,42)=(8,16)