{"id":309,"date":"2020-03-27T18:53:59","date_gmt":"2020-03-27T21:53:59","guid":{"rendered":"http:\/\/cinoto.com.br\/matematica\/?p=309"},"modified":"2020-03-27T18:53:59","modified_gmt":"2020-03-27T21:53:59","slug":"na-funcao-fxbx-sendo-0","status":"publish","type":"post","link":"http:\/\/cinoto.com.br\/matematica\/na-funcao-fxbx-sendo-0\/","title":{"rendered":"1) Na fun\u00e7\u00e3o f(x) = b^x, sendo 0 < b < 1, se f(1) + f(-1) = 10\/3, a \u00fanica afirmativa verdadeira sobre o valor de b \u00e9:"},"content":{"rendered":"\n<p>a)\u00a00 &lt; b &lt; 1\/9 \u00a0 \u00a0\u00a0b)\u00a02\/9 &lt; b &lt; 4\/9 \u00a0 \u00a0\u00a0c)\u00a08\/9 &lt; b &lt; 1 \u00a0 \u00a0\u00a0d)\u00a01 &lt; b &lt; 4 \u00a0 \u00a0\u00a0e)\u00a04 &lt; b &lt; 9<\/p>\n\n\n\n<!--more-->\n\n\n\n<h2 class=\"wp-block-heading\">Resolu\u00e7\u00e3o:<\/h2>\n\n\n\n<p>Se f(1) + f(-1) = 10\/3, podemos substituir os valores de f(1) e f(-1) em fun\u00e7\u00e3o de b, j\u00e1 que f(x) = b<sup>x<\/sup>:<\/p>\n\n\n\n<p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; f(x) = b<sup>x<\/sup>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;f(x) = b<sup>x<\/sup><\/p>\n\n\n\n<p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; f(1) = b<sup>1<\/sup>&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;f(-1) = b<sup>(-1)<\/sup><\/p>\n\n\n\n<p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; f(1) = b &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;f(-1) = 1 \/ b<\/p>\n\n\n\n<p>f(1) + f(-1) = 10\/3<\/p>\n\n\n\n<p>b + (1\/b) = 10\/3, tirando o m\u00ednimo,<\/p>\n\n\n\n<p>3b\u00b2\/3b + (3\/3b) = 10b\/3b, como b \u00e9 diferente de zero,<\/p>\n\n\n\n<p>3b\u00b2 + 3 = 10b<\/p>\n\n\n\n<p>3b\u00b2 &#8211; 10b+ 3 = 0, resolvendo a equa\u00e7\u00e3o do 2\u00ba grau,<\/p>\n\n\n\n<p>(3b &#8211; 1) . (b &#8211; 3) = 0<\/p>\n\n\n\n<p>b = 1\/3 &nbsp; ou &nbsp; b = 3<\/p>\n\n\n\n<p>Como, pelo problema, 0 &lt; b &lt; 1, b s\u00f3 pode ser 1\/3.&nbsp;Pelas alternativas, vemos que todas tem 9 como denominador. Ent\u00e3o vamos transformar 1\/3 para denominador 9:<\/p>\n\n\n\n<p>1\/3 = 3\/9<\/p>\n\n\n\n<p>Resposta: b) 2 \/ 9 &lt; b &lt; 4 \/ 9<\/p>\n","protected":false},"excerpt":{"rendered":"<p>a)\u00a00 &lt; b &lt; 1\/9 \u00a0 \u00a0\u00a0b)\u00a02\/9 &lt; b &lt; 4\/9 \u00a0 \u00a0\u00a0c)\u00a08\/9 &lt; b &lt; 1 \u00a0 \u00a0\u00a0d)\u00a01 &lt; b &lt; 4 \u00a0 \u00a0\u00a0e)\u00a04 &lt; b &lt; 9<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[11],"tags":[28],"class_list":["post-309","post","type-post","status-publish","format-standard","hentry","category-funcoes","tag-facil"],"_links":{"self":[{"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/posts\/309","targetHints":{"allow":["GET"]}}],"collection":[{"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/comments?post=309"}],"version-history":[{"count":1,"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/posts\/309\/revisions"}],"predecessor-version":[{"id":310,"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/posts\/309\/revisions\/310"}],"wp:attachment":[{"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/media?parent=309"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/categories?post=309"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/tags?post=309"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}