The linear approximation at x = 0 to f(x) = sqrt { 7 + 5 x } is y =

The linear approximation at x = 0 to f(x) = sqrt { 7 + 5 x } is y = ?

Linearization of f at a:

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

With a = 0:

L(x) = f'(0) *x + f(0)

f(0) = sqrt(7)

f'(x) = 5/(2 sqrt(7 + 5x))

So f'(0) = 5/(2 sqrt(7)) = 5 sqrt(7) /14

Then:
L(x) = 5 sqrt(7) x /14 + sqrt(7)
= sqrt(7) * ( 5x/14 + 1)

f'(a) ≈ [f(x) - f(a)]/(x-a)
y = f(x) ≈ f(a) + f'(a)(x-a)

f'(x) = 5/[2√(7+5x)]
f'(0) = 5/(2√7) = 5√7/14
f(0) = √7

y(x) = (5√7/14)x + √7
or
y(x) = 0.94491x + 2.64575

f(0.1) = 2.73861 (to 5 decimal places)
y(0.1) = 2.74024 (to 5 decimal places)
y(0.1) = f(0.1) to 3 significant digits.

f(-0.1) = 2.54951
y(-0.1) = 2.55126
y(-0.1) = f(-0.1) to 3 significant digits.