Question

Given the function how do I solve

Answer 1

Answer: (a) -7      (b) $\bold{\dfrac{4}{25}}$      (c) $\bold{-\dfrac{5}{2}}$    (d) 4

Step-by-step explanation:

A) f(x) = 2x - 3      when x is between -5 and -2 (including -5 and -2)

B) f(x) = x²            when x is between -2 and 2 (including 2)

C) $f(x)=-\dfrac{3}{2}x+\dfrac{7}{2}$             when x is between 2 and 5 (including 5)

a) Equation A includes x = -2

   f(x) = 2x - 3

   f(-2) = 2(-2) - 3

           =  -4    - 3

           =       -7

b) Equation B includes x = $-\dfrac{2}{5}$

   f(x) = x²

   $f\bigg(-\dfrac{2}{5}\bigg) = \bigg(-\dfrac{2}{5}\bigg)^2$

        $=\dfrac{4}{25}$

c) Equation C includes x = 4

   $f(x)=-\dfrac{3}{2}x+\dfrac{7}{2}$

   $f(4)=-\dfrac{3}{2}(4)+\dfrac{7}{2}$

       $=-\dfrac{12}{2}+\dfrac{7}{2}$

       $=-\dfrac{5}{2}$

d) Try each equation to see if x falls within the values given.

A) f(x) = 2x - 3

   -2.5 = 2x - 3

    0.5 = 2x

  0.25 = x    NOT VALID since x should be between -5 and -2

B) f(x) = x²

   -2.5 = x²

   $\sqrt{-2.5}=x$  NOT VALID since x cannot be an imaginary number

C) $f(x)=-\dfrac{3}{2}x+\dfrac{7}{2}$

    $-\dfrac{5}{2}=-\dfrac{3}{2}x+\dfrac{7}{2}$

    $-\dfrac{12}{2}=-\dfrac{3}{2}x$

    $-\dfrac{12}{2}\bigg(\dfrac{2}{3}\bigg)=x$

    $4 = x$        VALID since x is between 2 and 5