Evaluating functions
- Evaluate \(f(x)=2x^2-5x+1\) at \(x=-3\)
\[ f(-3) = 2(-3)^2-5(-3)+1 = 18+15+1 = 34 \]
- Evaluate \(g(a,b)=\frac{a^2+b}{2a}\) at \(a=4\) and \(b=10\)
\[ g(4,10) = \frac{4^2+10}{2(4)} = \frac{26}{8} = 3.25 \]
Lines & slope
- What are the \(x\)-intercept and \(y\)-intercept of the line \(y=4x-8\)?
\(x\)-intercept: set \(y=0\): \(0=4x-8 \Rightarrow x=2\), so \((2,0)\).
\(y\)-intercept: set \(x=0\): \(y=-8\), so \((0,-8)\).
- Find the equation of the line that passes through \((1,4)\) and \((5,-8)\).
\[ m=\frac{-8-4}{5-1}=\frac{-12}{4}=-3 \] \[ y-4=-3(x-1) \Rightarrow y=-3x+7 \]
- Find the equation of the line that passes through \((-2,3)\) and has slope \(\frac{1}{2}\).
\[ y-3=\frac{1}{2}(x+2) \Rightarrow y=0.5x+4 \]
- Between 2005 and 2015, a glacier’s area shrank from 120 km² to 80 km². Find the average rate of change and write a sentence interpreting it.
\[ m=\frac{80-120}{2015-2005}=\frac{-40}{10}=-4\ \frac{\text{km}^2}{\text{year}} \] Between 2005 and 2015, glacier area decreased by an average of 4 km² per year.
The exponential function
- Simplify: \(e^{2x}\cdot e^{-5x}\)
\[ e^{2x}\cdot e^{-5x} = e^{2x-5x}=e^{-3x} \]
- A population grows according to \(N(t)=50e^{0.08t}\). What is the population after 10 years?
\[ N(10)=50e^{0.08(10)}=50e^{0.8}\approx 50(2.2255)\approx 111 \]
- The logistic growth model is \[N_t=\frac{K}{1+\left[\frac{K-N_0}{N_0}\right]e^{-rt}}\] What does \(N_t\) approach as \(t \to \infty\) (assuming \(r>0\))?
As \(t \to \infty\), \(e^{-rt} \to 0\) (since \(r>0\)), so the bracketed term disappears: \[ N_t \to \frac{K}{1+0} = K \] The population approaches the carrying capacity \(K\).
Logarithms
- Solve for \(x\): \(\ln(x)=4\)
\[ x=e^4\approx 54.6 \]
- Solve for \(t\): \(200e^{0.05t}=500\)
\[ e^{0.05t}=2.5 \Rightarrow 0.05t=\ln(2.5)\approx 0.9163 \Rightarrow t\approx 18.33 \]
- Given \(\ln(2)\approx 0.693\) and \(\ln(3)\approx 1.099\), use log properties to find \(\ln(6)\).
\[ \ln(6) = \ln(2\cdot 3) = \ln(2)+\ln(3) \approx 0.693+1.099 = 1.792 \]