Question

Funcția $f: R -\ \textgreater \ R$,$f(x)=x^4-8x+1$ admite tangenta de ecuație $y=24x-47$. Să se determine coordonatele punctului de tangență.

Answer 1

Răspuns:

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Explicație pas cu pas:

Answer 2

$\it Fie\ A(x_0,\ y_0)\ punctul\ de\ tangen\c{\it t}\breve{a}\\ \\ Ecua\c{\it t}ia\ tangentei\ este:\ y-y_0=f'(x_0)(x-x_0)\ \ \ \ \ (1)$

$\it f'(x_0)=4x_0^3-8\ \ \ \ (2)\\ \\ (1),\ (2) \Rightarrow y=(4x_0^3-8)(x-x_0)+y_0=4x_0^3x-4x_0^4-8x+8x_0+y_0 \Rightarrow\\ \\ \Rightarrow y=(4x_0^3-8)x-4x_0^4+8x_0+y_0\ \ \ \ \ (3)\\ \\ \\ Dar,\ ecua\c{\it t}ia\ tangentei\ este:\ \ y=24x-47\ \ \ \ (4)\\ \\ (3),\ (4) \Rightarrow 4x_0^3-8=24|_{+8} \Rightarrow 4x_0^3=32|_{:4} \Rightarrow x_0^3=8 \Rightarrow x_0=2\ \ \ \ \ (5)$

$\it (3),\ (5) \Rightarrow y=24x-4\cdot16+16+y_0 \Rightarrow y=24x-64+16+y_0 \Rightarrow \\ \\ \Rightarrow y=24x-48+y_0\ \ \ \ (3')\\ \\ (3'),\ (4) \Rightarrow -48+y_0=-47 \Rightarrow y_0=48-47 \Rightarrow y_0=1\\ \\ \\ Deci,\ punctul\ cerut\ este\ A(2,\ 1)$