The zeros of a function \( f(x) \) are the values of \( x \) for which \( f(x) = 0 \). In other words, they are the x-coordinates of the points where the graph of the function intersects the x-axis.
Find the zero of the linear function:
\( f(x) = -2x + 4 \)
Set \( f(x) = 0 \) and solve for \( x \):
\( -2x + 4 = 0 \)
\( x = 2 \)
Find the zeros of the quadratic function:
\( f(x) = -2x^2 - 5x + 7 \)
Set \( f(x) = 0 \) and solve for \( x \):
\( -2x^2 - 5x + 7 = 0 \)
Factor or use the quadratic formula:
\( (-2x - 7)(x - 1) = 0 \)
Thus, the zeros are:
\( x = -\frac{7}{2} \) and \( x = 1 \)
Find the zeros of the trigonometric function:
\( f(x) = \sin(x) - \frac{1}{2} \)
Set \( f(x) = 0 \) and solve:
\( \sin(x) - \frac{1}{2} = 0 \)
\( \sin(x) = \frac{1}{2} \)
The solutions are:
\( x = \frac{\pi}{6} + 2k\pi \) and \( x = \frac{5\pi}{6} + 2k\pi \), where \( k \) is any integer.
Find the zero of the logarithmic function:
\( f(x) = \ln(x - 3) - 2 \)
Set \( f(x) = 0 \) and solve:
\( \ln(x - 3) - 2 = 0 \)
\( \ln(x - 3) = 2 \)
Convert to exponential form:
\( x - 3 = e^2 \)
Thus, the zero is:
\( x = 3 + e^2 \)
Find the zeros of the exponential function:
\( f(x) = e^{x^2 - 2} - 3 \)
Set \( f(x) = 0 \) and solve:
\( e^{x^2 - 2} - 3 = 0 \)
\( e^{x^2 - 2} = 3 \)
Take the natural logarithm:
\( x^2 - 2 = \ln(3) \)
Thus, the zeros are:
\( x = \sqrt{\ln(3) + 2} \) and \( x = -\sqrt{\ln(3) + 2} \)