f(x) = x^2 + 4x - 8
dy/dx = 2x+4
at x=2 , dy/dx = 8
f(2) = 4 + 8-8 = 4
tangent equation at x = 2
y-4 = 8(x-2)
y-4 = 8x-16
y = 8x -12
8x-12 = 2(x^2 + 4x-8)
8x - 12 = 2x^2 + 8x - 16
2x^2 = 4
x^2 = 2
x = ± √2
so what do you think?
1. Let f be the function given by f(x)=x^2+4x-8. The tangent line to the graph at x = 2 is used to approximate values of f . For what value(s) of x is the tangent line approximation twice that of f ?
I: -√2 II: 1 III: √2
Select one:
a. III only
b. I and III only
c. I only
d. II only
e. I and II only
2.Determine the values of x for which the linear approximation √(1+x) ≈1+(x/2) is accurate to within 0.11.
Select one:
a. -0.71 < x < 1.15
b. -1.71 < x < 0.15
c. -0.21 < x < 1.65
d. 0.29 < x < 2.15
1 answer