Hong Kong
Stage 4 - Stage 5

# Intersections of graphs by other methods II

## Interactive practice questions

Michael is going to use the point of intersection of graphs to find an approximation to the positive solution of $\log\left(x\right)-1=-x+5$log(x)1=x+5.

a

One of the graphs is $f\left(x\right)=\log\left(x\right)-1$f(x)=log(x)1. What function $g\left(x\right)$g(x) should the other graph be of?

b

The graph of $f\left(x\right)$f(x) has been provided. On the same set of axes, graph $g\left(x\right)=-x+5$g(x)=x+5.

c

Between which two values does the solution of the equation lie? There may be more than one solution.

$x=4$x=4 and $x=9$x=9

A

$x=2$x=2 and $x=5$x=5

B

$x=6$x=6 and $x=7$x=7

C

$x=5$x=5 and $x=6$x=6

D
Easy
1min

Consider the function $f\left(x\right)=\log_2\left(x+1\right)$f(x)=log2(x+1).

Easy
1min

Consider the functions $f\left(x\right)=x^3+2x^2$f(x)=x3+2x2 and $g\left(x\right)=x+4$g(x)=x+4.

Medium
2min

The populations of two competing species of animals is given by $f\left(t\right)=215t+3$f(t)=215t+3 and $g\left(t\right)=\frac{1}{t-3}$g(t)=1t3, where $f\left(t\right)$f(t) and $g\left(t\right)$g(t) represent the respective populations $t$t years from now. Jenny is looking to determine approximately when their populations will be equal.

Medium
4min