Question

URGENTE!!!

Caros, dada a função f: R R, definida por F(x) = 5x ao quadrado - 8x + 3, calcule;

a) f(-1)
b) f(0)
c) f (1/2)
d) f(raiz de 2)
e) x, de modo que f(x) = 0
f) x, de modo que f(x) = 7

Answer 1

Olá

Nos itens a, b, c, d ; basta substituir os valores dados no lugar do 'x'.

A)

f(x) = 5x² - 8x + 3
f(-1) = 5(-1)² - 8(-1) + 3
f(-1) = 5 + 8 + 3
f(-1) = 16


B)


f(x) = 5x² - 8x + 3
f(0) = 5(0)² - 8(0) + 3
f(0) = 0 + 0 + 3
f(0) = 3


C)

$\displaystyle \mathsf{f(x) = 5x^2 - 8x + 3} \\\\\\\mathsf{f( \frac{1}{2} ) = 5\cdot( \frac{1}{2} )^2 - 8\cdot( \frac{1}{2} ) + 3}\\\\\\\mathsf{f( \frac{1}{2} ) = 5\cdot( \frac{1}{4} ) - 4 + 3}\\\\\\\mathsf{f( \frac{1}{2} ) = \frac{5}{4} - 1}\\\\\\\boxed{\mathsf{f( \frac{1}{2} )= \frac{1}{4} }}$




D)

$\displaystyle \mathsf{f(x) = 5x^2 - 8x + 3} \\\\\\\mathsf{f( \sqrt{2} ) = 5\cdot( \sqrt{2} )^2 - 8\cdot( \sqrt{2} ) + 3}\\\\\\\mathsf{f(\sqrt{2} ) = 5\cdot( 2) - 8\sqrt{2} + 3}\\\\\\\mathsf{f( \sqrt{2} ) = 10-8\sqrt{2} +3}\\\\\\\mathsf{f( \sqrt{2} )= 13-8\sqrt{2} }\\\\\\\boxed{\mathsf{f(\sqrt{2})\approx 1,686}}$



E)

Nesse item, temos que encontrar um valor para 'x' de modo que, se substituirmos na f(x), a função irá zerar.
Para fazer isso, é bem simples, basta aplicar bhaskara.

$\displaystyle\mathsf{f(x)=5x^2-8x+3=0}\\\\\\\\\mathsf{\Delta=(-8)^2-4\cdot 5\cdot 3}\\\\\mathsf{\Delta=64-60}\\\\\mathsf{\Delta=4}\\\\\\\mathsf{ X=\frac{8\pm \sqrt{4}}{2\cdot 5} }\\\\\\\mathsf{ X_1=\frac{8+2}{10} }~~~~~ ~~\longrightarrow~\boxed{\mathsf{X_1=1}}~~~~~~~\longleftarrow\text{Resposta 1}\\\\\\\mathsf{ X_2=\frac{8-2}{10} }~~~~~ ~~\longrightarrow~\boxed{\mathsf{X_2= \frac{3}{5} }}~~~~~~~\longleftarrow\text{Resposta 2}$



F)

Nesse item, temos que encontrar um ou mais valores para 'x' de modo que, quando substituirmos, resulte em 7.
Para isso, temos que igualar a função a 7

$\mathsf{5x^2-8x+3=7}\\\\\text{Passa o 7 para o outro lado}\\\\\mathsf{5x^2-8x+3-7=0}\\\\\mathsf{5x^2-8x-4=0}$

Agora, é o mesmo procedimento do item anterior, encontrar os valores de 'x' por bhaskara.

$\displaystyle\mathsf{5x^2-8x-4=0}\\\\\\\\\mathsf{\Delta=(-8)^2-4\cdot 5\cdot (-4)}\\\\\mathsf{\Delta=64+80}\\\\\mathsf{\Delta=144}\\\\\\\mathsf{ X=\frac{8\pm \sqrt{144}}{2\cdot 5} }\\\\\\\mathsf{ X_1=\frac{8+12}{10} }~~~~~ ~~\longrightarrow~\boxed{\mathsf{X_1=2}}~~~~~~~\longleftarrow\text{Resposta 1}\\\\\\\mathsf{ X_2=\frac{8-12}{10} }~~~~~ ~~\longrightarrow~\boxed{\mathsf{X_2= -\frac{2}{5} }}~~~~~~~\longleftarrow\text{Resposta 2}$