{"id":983,"date":"2015-01-23T10:38:30","date_gmt":"2015-01-23T09:38:30","guid":{"rendered":"http:\/\/www.unimath.fr\/?p=983"},"modified":"2016-05-28T15:28:13","modified_gmt":"2016-05-28T14:28:13","slug":"dm-guide-lecture-graphique-primitive","status":"publish","type":"post","link":"http:\/\/www.unimath.fr\/?p=983","title":{"rendered":"DM Guid\u00e9 Fonction exponentielle n\u00b03 – Lecture graphique – Primitive"},"content":{"rendered":"

Sujet : \u00a0Cliquer ici<\/a>
\nCorrection en fin de page<\/p>\n

1. Signe de f'(x)<\/span>
\n[peekaboo_link name=\u00a0\u00bbAide1″]Aide[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbAide1″] Les variations de la fonction donnent le signe de la d\u00e9riv\u00e9e[\/peekaboo_content]
\n2. a. Courbe de f’ et courbe de F ?
\n<\/span>[peekaboo_link name=\u00a0\u00bbAide2″]R\u00e9ponse courbe de f'[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbAide2″]D’apr\u00e8s le tableau de variations de f, f’ est positive puis n\u00e9gative donc la courbe de f’ est C2[\/peekaboo_content]
\nb. Aide pour a :
\n<\/span>[peekaboo_link name=\u00a0\u00bbAide3″]Aide 1[\/peekaboo_link]
\n[peekaboo_content name=\u00a0\u00bbAide3″]f admet un maximum en a donc f'(a)=0[\/peekaboo_content]
\n[peekaboo_link name=\u00a0\u00bbAide4″]Aide 2[\/peekaboo_link]
\n[peekaboo_content name=\u00a0\u00bbAide4″]Regarder o\u00f9 la courbe de f’, (C2) coupe l’axe des abscisses[\/peekaboo_content]
\nb. Aide pour b :
\n<\/span>[peekaboo_link name=\u00a0\u00bbAide5″]Aide 1[\/peekaboo_link]
\n[peekaboo_content name=\u00a0\u00bbAide5″]Utiliser F’ et se rappeler que F'(x) = f(x). Une valeur de f(x) est connue dans le tableau. Laquelle ?[\/peekaboo_content]
\n[peekaboo_link name=\u00a0\u00bbAide6″]Aide 2[\/peekaboo_link]
\n[peekaboo_content name=\u00a0\u00bbAide6″]On a f(a) = b, donc F'(a) = b. Que repr\u00e9sente F'(a) pour la courbe de F’ (C1) ?[\/peekaboo_content]
\n[peekaboo_link name=\u00a0\u00bbAide7′]Aide 3[\/peekaboo_link]
\n[peekaboo_content name=\u00a0\u00bbAide7″]F'(a) est le coefficient directeur de la tangente \u00e0 la courbe de F, (C1) au point d’abscisse a. Quel est le signe de ce coefficient directeur d’apr\u00e8s la courbe ?[\/peekaboo_content]
\n3. Information donn\u00e9e par le point A ?<\/span>
\n[peekaboo_link name=\u00a0\u00bbAide8″]R\u00e9ponse[\/peekaboo_link]
\n[peekaboo_content name=\u00a0\u00bbAide8″]A(0,2) appartient \u00e0 la courbe de F donc F(0) = 2[\/peekaboo_content]
\nInformation donn\u00e9e par le point B ?<\/span>
\n[peekaboo_link name=\u00a0\u00bbAide9″]R\u00e9ponse[\/peekaboo_link]
\n[peekaboo_content name=\u00a0\u00bbAide9″]B(0,1\/2) appartient \u00e0 la courbe de f’ donc f'(0) = 1\/2[\/peekaboo_content]
\nAide pour f'(x) :<\/span>
\n[peekaboo_link name=\u00a0\u00bbAide10″]Aide[\/peekaboo_link]
\n[peekaboo_content name=\u00a0\u00bbAide10″]Calculer f'(x) puis trouver k sachant que f'(0) = 1\/2[\/peekaboo_content]
\n[peekaboo_link name=\u00a0\u00bbAide11″]R\u00e9ponse[\/peekaboo_link]
\n[peekaboo_content name=\u00a0\u00bbAide11″]k = -1[\/peekaboo_content]
\nAide pour F(x) : (primitive de f)<\/span>
\n[peekaboo_link name=\u00a0\u00bbAide12″]Aide[\/peekaboo_link]
\n[peekaboo_content name=\u00a0\u00bbAide12″]Une primitive de - e^{\\frac{1}{2}x} est -2 e^{\\frac{1}{2}x} [\/peekaboo_content]
\n[peekaboo_link name=\u00a0\u00bbAide13″]R\u00e9ponse[\/peekaboo_link]
\n[peekaboo_content name=\u00a0\u00bbAide13″]F(x)=-2 e^{\\frac{1}{2}x} + \\dfrac{x^2}{2}+2x [\/peekaboo_content]
\nPour trouver a :<\/span>
\n[peekaboo_link name=\u00a0\u00bbAide14″]Aide[\/peekaboo_link]
\n[peekaboo_content name=\u00a0\u00bbAide14″]R\u00e9soudre f'(x) = 0[\/peekaboo_content]
\n[peekaboo_link name=\u00a0\u00bbAide15″]R\u00e9ponse[\/peekaboo_link]
\n[peekaboo_content name=\u00a0\u00bbAide15″]a = 2 ln2[\/peekaboo_content]
\nPour trouver b :<\/span>
\n[peekaboo_link name=\u00a0\u00bbAide16″]Aide[\/peekaboo_link]
\n[peekaboo_content name=\u00a0\u00bbAide16″]Rappel : b = f(a)[\/peekaboo_content]
\n[peekaboo_link name=\u00a0\u00bbAide17″]R\u00e9ponse[\/peekaboo_link]
\n[peekaboo_content name=\u00a0\u00bbAide17″]b= f(2 ln2) = 2 ln2[\/peekaboo_content]<\/p>\n

 <\/p>\n

Correction : <\/span> cliquer ici<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Sujet : \u00a0Cliquer ici Correction en fin de page 1. Signe de f'(x) [peekaboo_link name=\u00a0\u00bbAide1″]Aide[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbAide1″] Les variations de la fonction donnent le signe de la d\u00e9riv\u00e9e[\/peekaboo_content] 2. a. Courbe de f’ et courbe de F ? [peekaboo_link name=\u00a0\u00bbAide2″]R\u00e9ponse courbe de f'[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbAide2″]D’apr\u00e8s le tableau de variations de f, f’ est positive puis n\u00e9gative donc […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":[],"categories":[1],"tags":[],"_links":{"self":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts\/983"}],"collection":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=983"}],"version-history":[{"count":39,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts\/983\/revisions"}],"predecessor-version":[{"id":2325,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts\/983\/revisions\/2325"}],"wp:attachment":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=983"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=983"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=983"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}