f(x)=x³
First find the derivative (slope) of f(x):
f'(x)=3x² using standard rules of power.
Slope at x=4
f'(4)=3*4²=48
The value of the function at x=4 is
f(4)=4³=64
This the tangent sought is tangent to the curve at P(4,192):
L: y=f'(4)x+192=48x+192
To find an approximation at 3.8³, we use linearization which uses the slope at x=4 (close to 3.8) equal to f'(4)=48.
The approximate value of f(3.8) is
f(x-b)≅f(x)-b*f'(x).
Putting x=4, b=0.2, x-b=3.8
f(3.8)≅f(4)-0.2*48
=64-0.2*48
=64-9.6
=54.4
Exact value=3*3.8³=54.87
Error=(54.4-54.87)/54.87=0.9%
If f(x)=x^3. Yhe equation of the tangent line to f(x) at x=4 is y=?
Using this, we find our approximation for 3.8^3 is ?
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