Search results
Results from the WOW.Com Content Network
The exponential function calculator will help you solve commonly used forms of the exponential function, given some known points on the line. The calculator will also let you evaluate any exponential function with any parameters.
For the following exercises, find the formula for an exponential function that passes through the two points given.
Use this step-by-step Exponential Function Calculator, to find the function that describe the exponential function that passes through two given points.
Find an exponential function that passes through the points \((−2,6)\) and \((2,1)\). Solution Because we don’t have the initial value, we substitute both points into an equation of the form \(f(x)=ab^x\), and then solve the system for \(a\) and \(b\).
An exponential function represents the relationship between an input and output, where we use repeated multiplication on an initial value to get the output for any given input. Exponential functions can grow or decay very quickly.
A function that models exponential growth grows by a rate proportional to the current amount. For any real number x and any positive real numbers a and b such that b ≠1 b ≠ 1, an exponential growth function has the form. f(x) =abx f (x) = a b x. where. a is the initial or starting value of the function.
Write the equation for the exponential function passing through each pair of points below. Round any decimals to the nearest thousandth.
Given the two points [latex]\left(1,3\right)[/latex] and [latex]\left(2,4.5\right)[/latex], find the equation of the exponential function that passes through these two points. Solution
At x=1, f (x)=a ... in other words it passes through (1,a) It is an Injective (one-to-one) function. Its Domain is the Real Numbers: Its Range is the Positive Real Numbers: (0, +∞) Inverse. ax is the inverse function of loga(x) (the Logarithmic Function) So the Exponential Function can be "reversed" by the Logarithmic Function.
Find an exponential function that passes through the points \((−2,6)\) and \((2,1)\). Solution Because we don’t have the initial value, we substitute both points into an equation of the form \(f(x)=ab^x\), and then solve the system for \(a\) and \(b\).