Use linear approximation, i.e. the tangent line, to approximate 6.7^3 as follows: Let f(x)=x^3. The equation of the tangent line to f(x) at x=7 can be written in the form y=mx+b

Question

Use linear approximation, i.e. the tangent line, to approximate 6.7^3 as follows:

Let f(x)=x^3. The equation of the tangent line to f(x) at x=7 can be written in the form y=mx+b

where m is:  

and where b is:  

Using this, we find our approximation for 6.7^3 is

Answer 1

The slope of the line tangent to f(x) = x³ is the derivative function. The derivative function, using the power rule, is f '(x) = 3x² so at x = 7, the slope is f '(7) = 3(7²) = 3(49) = 147.

The function value for x = 7 is 7³ = 343 so (7, 343) is the point on the curve we are being asked to find the equation of the tangent line for. We can use the point-slope form:

y - 343 = 147(x - 7)

y - 343 = 147x - 1029

y = 147x - 686 or, using function notation:

L(x) = 147x - 686

So m = 147 and b = -686

So the linear approximation of 6.7^3 is:

L(6.7) = 147(6.7) - 686 = 298.9

Answer 2

Hi Chelle M.,

Solve this one bit-by-bit. First, take the first derivative f ' (x) = ? (You should know how to take for a polynomial function!). Plug in the value for x=7, and obtain the slope m. Next, calculate 7^3 (by hand, whatever), that gives you a point (7, 7^3) which is on that tangent line! (So you can calculate b .)

So, you are supposed to follow this tangent line down to x=6.7, and read the y-value. Piece of cake, right?

But this is vastly less efficient than just calculating it!

Now let's test your extrapolation powers, using powers. Consider the number 6^100000. (It's my current favorite integer!) How many digits does it have (hint: if you expect Excel to purr and show you directly, you are sadly mistaken!). How do you think you could manipulate such a number (yes, do actual math on it!)?

-- Cheers, --Mr. d.