If f(x)=2x^2-4x+7, find f'(2)

Question

If f(x)=2x²-4x+7, find f'(2) _____

Use this to find the linear approximation to  at .

The equation of this linear approximation is

L(x)= _____

Use  to approximate . (Compute the actual value of .)

f(2.2) ≈ _____

Compare this with the actual value of f(2.2) = _____

Answer 1

f(x) = 2x^2 -4x + 7

f'(x) = 4x -4

f'(2) = 4(2)-4 = 4

L(x) =f(x) +f'(x)(x-a)

f(x) = 2x^2 -4x + 7

f'(x) = 4x-4

L(x) = 2x^2 -4x + 7+(4x-4)(x-a)

tangent line at x=2, at (2, f(x)) at (2,4) is f'(x) = (y-4)/(x-2) = 4x-4= 4(2)-4 = 4

cross multiply

y-4 = 4x-8

y = 4x-4 is the tangent line whose slope = 4 = f'(x) = the derivative

the equation describes an upward opening parabola with vertex (1,5)

at (2,7) the tangent line has slope = 4 = f'(2)

for x-2.2

y = 4(2.2)-4 = 8.8-4 = 4.8 = a close "approximation" to the slope at x=2.2 = exactly the derivative f'(2.2) which = 4x -4 = 4(2.2) -4 = 8.8-4 = 4.8

f'(x) = 4x-4

f(2) = 4

f(2.2) = 4(2.2)-4 = 8.8-4 = 4.8