Locating the zeros of a function is an important skill. Functions switch signs at zeros which can reflect a lot of important information - especially when the function is a derivative where switching signs signals switching behavior.
We don’t have many ways of locating zeros for complicated functions.
\(\blacktriangleright \) Zeros of linear functions are easy to locate - just solve the equation for the variable.
\(\blacktriangleright \) Basic exponential functions don’t have zeros. Core logarthmic functions have a zero when the inside of the formula equals \(1\). For more complicated exponential and logarithmic functions, we can use transformations and the algebra rules.
\(\blacktriangleright \) Sine and cosine have repeating zeros. So, we can identify them for one period and then use transformations to locate the others.
\(\blacktriangleright \) We have several methods for quadratic functions. We can complete the square and use the quadratic formula.
Most functions are not these.
Generally speaking, we have only two ways to solve equations, which gives us two methods to finding zeros for a function.
- (a)
- If you know what to do, then do it.
- (b)
- If you con’t know what to do, then get everything on one side and \(0\) on the other side of the equation and factor. Use the Zero Product Property.
The Zero Product Property is one of the few simple facts we know about the arithmetic of the real numbers.
Let \(a\) and \(b\) be real numbers.
If \(a \cdot b = 0\), then either \(a=0\) or \(b=0\).
That is one of our best tools for locating zeros of functions (like critical numbers).
Solve \(k^2 - k - 20 = 0\)
explanation
Rewrite \(k^2 - k - 20\) as a product.
If \(k^2 - k - 20 = (factor) \cdot (factor)\), then the factors will have to be linear or constants. If one of the factors was a constant, then we would have recognized a constant factor in all three coefficients of the terms of the expression. We can look beyond constant factors for this quadratic.
Therefore, we can begin with a template involving two linear factors, like \(k^2 - k - 20 = (k + ?) \cdot (k + ?)\)
When we multiply the factors out, the product of the two constant terms will have to be \(\answer {-20}\).
Possibilities:
- \(-20 \cdot 1\)
- \(20 \cdot \answer {-1}\)
- \(-10 \cdot 2\)
- \(10 \cdot \answer {-2}\)
- \(-5 \cdot 4\)
- \(5 \cdot \answer {-4}\)
Only one of these options gives a linear term of \(-k\).
\(k^2 - k - 20 = (k + 4) \cdot (k + (-5))\)
\(k^2 - k - 20 = (k + 4) \cdot (k - 5)\)
Now we can replace the original equation with an equivalent one:
\((k + 4) \left ( \answer {k - 5} \right ) = 0\)
Applying the zero product property tells us that either \(k + 4 = 0\) or \(k - 5 = 0\).
The solution set is \(\{ -4, 5 \}\)
Solve \(3 \, t \, e^t - (t-1) \, e^t = 0\)
explanation
This expression is a difference of two terms and each term has \(e^t\) as a factor. Therefore, we will apply the Distributive Property.
\(e^t \, (3t - (t-1))= 0\)
We can simplify the parentheses.
\(e^t \left ( \answer {2t + 1} \right )= 0\)
The zero product property tells us that either \(\answer {e^t} = 0\) or \(2t + 1 = 0\).
The function \(f(t) = e^t\) has no zeros, since it is an exponential function.
\(2t + 1 = 0\), when \(t = -\frac {1}{2}\)
We have one solution: \(\left \{ -\frac {1}{2} \right \}\)
We know two methods for solving equations.
\(\blacktriangleright \) If you are familiar with the function’s properties and have an idea of what to do, then go do that.
\(\blacktriangleright \) If you do not know what to do, then get everything on one side of the equation with zero on the other side, then factor.
If you don’t know what to do, then you know what to do. Factor and use the zero product property.
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more examples can be found by following this link
More Examples of Function Zeros