Homework Help Forum: calculus

Posted by Jenny on Thursday, November 5, 2009 at 4:09pm.

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 ?

• calculus - MathMate, Thursday, November 5, 2009 at 4:50pm

  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%

  
  

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