Questions

1 / 10

Time

Score

Linearization Of Logarithmic Functions

Determine the linear approximation of at a=0.

$L\left(x\right)=\log$ ₂ $\left(-1\right)-\frac{x}{\ln\left(2\right)_{ }}$

$L\left(x\right)=\log$

$L\left(x\right)=\log2$

Determine the linear approximation of at a=0.

$L\left(x\right)=\frac{x}{\ln\left(2\right)}$

$L\left(x\right)=\frac{1}{\ln\left(2\right)}$

$L\left(x\right)=\frac{x}{\ln\left(2\right)}$

Determine the linear approximation of at a=1.

$L\left(x\right)=1+\frac{x-1}{2\ln\left(2\right)}$

$L\left(x\right)=1+\frac{x+1}{2\ln\left(2\right)}$

$L\left(x\right)=1+\frac{x+1}{\ln\left(2\right)}$

Determine the linear approximation of at a=-1.

$L\left(x\right)=\log_3\left(-2\right)-\frac{x+1}{2\ln\left(3\right)}$

$L\left(x\right)=\log_3\left(-2\right)-\frac{x}{2\ln\left(3\right)}$

$L\left(x\right)=\log_3\left(-2\right)-\frac{x}{2\ln\left(1\right)}$

Determine the linear approximation of at a=1.

$L\left(x\right)=\frac{x+1}{\ln\left(5\right)}$

$L\left(x\right)=\frac{x-1}{\ln\left(5\right)}$

$L\left(x\right)=\frac{x}{\ln\left(5\right)}$

Determine the linear approximation of at a=0.

undefined

defined

$L\left(x\right)=0$

Determine the linear approximation of at a=0.

undefined

defined

$L\left(x\right)=2x$

Determine the linear approximation of at a=1.

defined

undefined

$L\left(x\right)=3$

Determine the linear approximation of at a=1.

$L\left(x\right)=\frac{x-1}{\ln\left(4\right)}$

$L\left(x\right)=\frac{x+1}{\ln\left(4\right)}$

$L\left(x\right)=\frac{x}{\ln\left(4\right)}$

Determine the linear approximation of at a=1.

$L\left(x\right)=\frac{x+1}{\ln\left(4\right)}$

$L\left(x\right)=\frac{x-1}{\ln\left(4\right)}$

$L\left(x\right)=\frac{x}{\ln\left(4\right)}$

Questions

Time

Score

Linearization Of Logarithmic Functions

Determine the linear approximation of at a=0.

﻿L(x)=log⁡L\left(x\right)=\logL(x)=log﻿ 2﻿(−1)−xln⁡(2)\left(-1\right)-\frac{x}{\ln\left(2\right)_{ }}(−1)−ln(2)​x​﻿

﻿L(x)=log⁡L\left(x\right)=\logL(x)=log﻿

﻿L(x)=log⁡2L\left(x\right)=\log2L(x)=log2﻿

Determine the linear approximation of at a=0.

﻿L(x)=xln⁡(2)L\left(x\right)=\frac{x}{\ln\left(2\right)}L(x)=ln(2)x​﻿

﻿L(x)=1ln⁡(2)L\left(x\right)=\frac{1}{\ln\left(2\right)}L(x)=ln(2)1​﻿

﻿L(x)=xln⁡(2)L\left(x\right)=\frac{x}{\ln\left(2\right)}L(x)=ln(2)x​﻿

Determine the linear approximation of at a=1.

﻿L(x)=1+x−12ln⁡(2)L\left(x\right)=1+\frac{x-1}{2\ln\left(2\right)}L(x)=1+2ln(2)x−1​﻿

﻿L(x)=1+x+12ln⁡(2)L\left(x\right)=1+\frac{x+1}{2\ln\left(2\right)}L(x)=1+2ln(2)x+1​﻿

﻿L(x)=1+x+1ln⁡(2)L\left(x\right)=1+\frac{x+1}{\ln\left(2\right)}L(x)=1+ln(2)x+1​﻿

Determine the linear approximation of at a=-1.

﻿L(x)=log⁡3(−2)−x+12ln⁡(3)L\left(x\right)=\log_3\left(-2\right)-\frac{x+1}{2\ln\left(3\right)}L(x)=log3​(−2)−2ln(3)x+1​﻿

﻿L(x)=log⁡3(−2)−x2ln⁡(3)L\left(x\right)=\log_3\left(-2\right)-\frac{x}{2\ln\left(3\right)}L(x)=log3​(−2)−2ln(3)x​﻿

﻿L(x)=log⁡3(−2)−x2ln⁡(1)L\left(x\right)=\log_3\left(-2\right)-\frac{x}{2\ln\left(1\right)}L(x)=log3​(−2)−2ln(1)x​﻿

Determine the linear approximation of at a=1.

﻿L(x)=x+1ln⁡(5)L\left(x\right)=\frac{x+1}{\ln\left(5\right)}L(x)=ln(5)x+1​﻿

﻿L(x)=x−1ln⁡(5)L\left(x\right)=\frac{x-1}{\ln\left(5\right)}L(x)=ln(5)x−1​﻿

﻿L(x)=xln⁡(5)L\left(x\right)=\frac{x}{\ln\left(5\right)}L(x)=ln(5)x​﻿

Determine the linear approximation of at a=0.

undefined

defined

﻿L(x)=0L\left(x\right)=0L(x)=0﻿

Determine the linear approximation of at a=0.

undefined

defined

﻿L(x)=2xL\left(x\right)=2xL(x)=2x﻿

Determine the linear approximation of at a=1.

defined

undefined

﻿L(x)=3L\left(x\right)=3L(x)=3﻿

Determine the linear approximation of at a=1.

﻿L(x)=x−1ln⁡(4)L\left(x\right)=\frac{x-1}{\ln\left(4\right)}L(x)=ln(4)x−1​﻿

﻿L(x)=x+1ln⁡(4)L\left(x\right)=\frac{x+1}{\ln\left(4\right)}L(x)=ln(4)x+1​﻿

﻿L(x)=xln⁡(4)L\left(x\right)=\frac{x}{\ln\left(4\right)}L(x)=ln(4)x​﻿

Determine the linear approximation of at a=1.

﻿L(x)=x+1ln⁡(4)L\left(x\right)=\frac{x+1}{\ln\left(4\right)}L(x)=ln(4)x+1​﻿

﻿L(x)=x−1ln⁡(4)L\left(x\right)=\frac{x-1}{\ln\left(4\right)}L(x)=ln(4)x−1​﻿

﻿L(x)=xln⁡(4)L\left(x\right)=\frac{x}{\ln\left(4\right)}L(x)=ln(4)x​﻿

Incorrect

Great!

$L\left(x\right)=\log$ ₂ $\left(-1\right)-\frac{x}{\ln\left(2\right)_{ }}$

$L\left(x\right)=\log$

$L\left(x\right)=\log2$

$L\left(x\right)=\frac{x}{\ln\left(2\right)}$

$L\left(x\right)=\frac{1}{\ln\left(2\right)}$

$L\left(x\right)=\frac{x}{\ln\left(2\right)}$

$L\left(x\right)=1+\frac{x-1}{2\ln\left(2\right)}$

$L\left(x\right)=1+\frac{x+1}{2\ln\left(2\right)}$

$L\left(x\right)=1+\frac{x+1}{\ln\left(2\right)}$

$L\left(x\right)=\log_3\left(-2\right)-\frac{x+1}{2\ln\left(3\right)}$

$L\left(x\right)=\log_3\left(-2\right)-\frac{x}{2\ln\left(3\right)}$

$L\left(x\right)=\log_3\left(-2\right)-\frac{x}{2\ln\left(1\right)}$

$L\left(x\right)=\frac{x+1}{\ln\left(5\right)}$

$L\left(x\right)=\frac{x-1}{\ln\left(5\right)}$

$L\left(x\right)=\frac{x}{\ln\left(5\right)}$

$L\left(x\right)=0$

$L\left(x\right)=2x$

$L\left(x\right)=3$

$L\left(x\right)=\frac{x-1}{\ln\left(4\right)}$

$L\left(x\right)=\frac{x+1}{\ln\left(4\right)}$

$L\left(x\right)=\frac{x}{\ln\left(4\right)}$

$L\left(x\right)=\frac{x+1}{\ln\left(4\right)}$

$L\left(x\right)=\frac{x-1}{\ln\left(4\right)}$

$L\left(x\right)=\frac{x}{\ln\left(4\right)}$