W Woodstalk New member Joined Jun 19, 2019 Messages 1 Jun 19, 2019 #1 This is the problem: Use a linear approximation to estimate (1.02)^3 I get how to do linear approximation with a function but I don't get this. Any help would be appreciated!
This is the problem: Use a linear approximation to estimate (1.02)^3 I get how to do linear approximation with a function but I don't get this. Any help would be appreciated!
MarkFL Super Moderator Staff member Joined Nov 24, 2012 Messages 3,021 Jun 19, 2019 #2 Hello, and welcome to FMH! We are being asked to estimate the cube of a quantity, so let's let: [MATH]f(x)=x^3[/MATH] Now, we can use the approximation: [MATH]\frac{\Delta f}{\Delta x}\approx\frac{df}{dx}[/MATH] Hence: [MATH]f(x+\Delta x)\approx\frac{df}{dx}\Delta x+f(x)[/MATH] Now, using the function we defined, with \(x=1\) and \(\Delta x=0.02\), we may state: [MATH](1+0.02)^3\approx3(1)^2(0.02)+1^3[/MATH] Or [MATH](1.02)^3\approx1.06[/MATH] Does that make sense?
Hello, and welcome to FMH! We are being asked to estimate the cube of a quantity, so let's let: [MATH]f(x)=x^3[/MATH] Now, we can use the approximation: [MATH]\frac{\Delta f}{\Delta x}\approx\frac{df}{dx}[/MATH] Hence: [MATH]f(x+\Delta x)\approx\frac{df}{dx}\Delta x+f(x)[/MATH] Now, using the function we defined, with \(x=1\) and \(\Delta x=0.02\), we may state: [MATH](1+0.02)^3\approx3(1)^2(0.02)+1^3[/MATH] Or [MATH](1.02)^3\approx1.06[/MATH] Does that make sense?