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avatar+45 

1.) Let $f(x)=\left\lfloor\left(-\frac58\right)^x\right\rfloor$ be a function that is defined for all values of $x$ in $[0,\infty)$ such that $f(x)$ is a real number. How many distinct values exist in the range of $f(x)$?
 

2.) Let \[f(x) = \left\{\begin{array}{cl}ax+3 & \text{ if }x>0, \\ab & \text{ if }x=0, \\bx+c & \text{ if }x<0.\end{array}\right.\]If $f(2)=5$, $f(0)=5$, and $f(-2)=-10$, and $a$, $b$, and $c$ are nonnegative integers, then what is $a+b+c$?
 

 

WITH LATEX:

 

1.) Let \(f(x)=\left\lfloor\left(-\frac58\right)^x\right\rfloor\) be a function that is defined for all values of \(x\) in \([0,\infty)\) such that \(f(x)\) is a real number. How many distinct values exist in the range of \(f(x)\)?

 

 

 

2.) Let

\(f(x) = \left\{\begin{array}{cl}ax+3 & \text{ if }x>0, \\ab & \text{ if }x=0, \\bx+c & \text{ if }x<0.\end{array}\right.\)

If \(f(2)=5\), \(f(0)=5\), and \(f(-2)=-10\), and \(a\), \(b\), and \(c\) are nonnegative integers, then what is \(a+b+c\)?

 Apr 26, 2019
 #1
avatar+106993 
+3

2) 

 

f(2)=2a+3=5

2a=2

a=1

 

f(0)=ab=5

1*b=5

b=5

 

f(-2)=b*-2+c=-10

-2b+c=-10

-2*5+c=-10

-10+c=-10

c=0

 

a+b+c =       well you can add up single digit numbers I expect. 

 Apr 26, 2019
 #3
avatar+45 
+1

Thank you so much!!! Both the answers were correct.

lolzforlife  Apr 27, 2019
 #2
avatar+106993 
+3

1.) Let \(f(x)=\left\lfloor\left(-\frac58\right)^x\right\rfloor\)   be a function that is defined for all values of x in \([0,\infty)\) such that f(x) is a real number. How many distinct values exist in the range of f(x)?

 

The real solutions     (-5/8)^x    are  always going to be between -1 and 1 not inclusive   (when x is positive)

so the floor function can only be  -2, -1  and 0

 

Hence you can answer the actual question.

 Apr 26, 2019

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