Linear approximation practice question 15
A student uses the linear approximation of the function below at to estimate , as shown. Find the first line that is wrong.
Check each line against the definition: the tangent line at 8 has the value and the slope at .
A check on the final answer: is about 19.5, nowhere near 8.3.
Put into the formula for that the student wrote. Does it give , as a tangent line at 8 must?
Theory UBC Math 100 §3.3.3CLP-1 §3.4Key results
Corrected working
Strategy
Tangent line at 8, where the cube root and its derivative are exact, then a step of 0.3 along it.
Working
Value and slope at the base point.
The slope multiplies the distance from the base point.
The true value is 2.0247…, so the estimate is good to three decimal places.
Answer
The answer is the cube root of 8.3 is about 81/40, that is, 2.025.
Check
2.025³ ≈ 8.304, just above 8.3, while the student’s 2.69 cubes to about 19.5.
Where this goes wrong
Multiplying the slope by x instead of by the step x − 8. The line then misses the point (8, 2) it is meant to pass through, and the estimate is far off.
The idea behind it
The linear approximation lesson, a figure to move, in a few steps
Leans on
Equation of a tangent line, from The definition of the derivative
Try a similar question
Use a linear approximation of at to estimate .Saved in this browser only.