How to Find Zeros of a Function

How to find the zeros of functions; tutorial with examples and detailed solutions. The zeros of a function f are found by solving the equation f(x) = 0.

Example 1: Find the zero of the linear function f is given by

f(x) = -2 x + 4

Solution to Example 1

To find the zeros of function f, solve the equation

f(x) = -2x + 4 = 0

Hence the zero of f is give by

x = 2

Example 2: Find the zeros of the quadratic function f is given by

f(x) = -2 x 2 - 5 x + 7

Solution to Example 2

Solve f(x) = 0

f(x) = -2 x 2 - 5 x + 7 = 0

Factor the expression -2 x 2 - 6 x + 8

(-2x - 7)(x - 1) = 0

and solve for x

x = -7 / 2 and x = 1

The graph of function f is shown below. The zeros of a function are the x coordinates of the x intercepts of the graph of f.

find zeros, example 2

Example 3: Find the zeros of the sine function f is given by

f(x) = sin(x) - 1 / 2

Solution to Example 3

Solve f(x) = 0

sin (x) - 1 / 2 = 0

Rewrite as follows

sin (x) = 1 / 2

The above equation is a trigonometric equation and has an infinite number of solutions given by

x = Pi / 6 + 2 k Pi and x = 5 Pi / 6 + 2 k Pi where k is any integer taking the values 0 , 1, -1, 2, -2 ...

The graph of f is shown below. The number of zeros of function f defined by f(x) = sin(x) - 1 / 2 are is infinite simply because function f is periodic.

find zeros, example 3

Example 4: Find the zeros of the logarithmic function f is given by

f(x) = ln (x - 3) - 2

Solution to Example 4

Solve f(x) = 0

ln (x - 3) - 2 = 0

Rewrite as follows

ln (x - 3) = 2

Rewrite the above equation changing it from logarithmic to exponential form

x - 3 = e 2

and solve to find one zero

x = 3 + e 2

Example 5: Find the zeros of the exponential function f is given by

f(x) = ex2 - 2 - 3

Solution to Example 4

Solve f(x) = 0

ex2 - 2 - 3 = 0

Rewrite the above equation as follows

ex2 - 2 = 3

Rewrite the above equation changing it from exponential to logarithmic form

x2 - 2 = ln (3)

Solve the above equation to find two zeros of f

x = square root [ln (3) + 2] and x = - square root [ln (3) + 2]

More tutorials on functions.

Applications, Graphs, Domain and Range of Functions


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Updated: 2 April 2013

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