given the function sqrtx^2+24 and linear approximation of the?

Question

function 9/5+x/5 when x0=1 approximate sqrt28

original function is f(x)=sqrtx^2+24
i found the linear approximation which is 9/5+x/5
at the point when x=1
my question is: use the linear approximation above to approximate sqrt28
hope this is clearer..thanx

Answer 1

y = sqrt(x^2+24) = (x^2+24)^1/2
dy/dx = x(x^2+24)^-1/2
y(1) = 5
y'(1) = 1/5
y - 5 = (1/5)(x - 1) = x/5 - 1/5

The approximation around x = 1 is
y = x/5 + 24/5, not 9/5 + x/5

to approximate √28,
28 - 24 = 4, √4 = 2
y ≈ 2/5 + 24/5
y ≈ 5.2

√28 ≈ 5.2915, so the error is about 1.7%

Answer 2

Something is definitely wrong here, please post the question again and don't try to write everything in the title or please add additional details by properly writing which function and the proper value of x= to what derised value, 5 or 6, because you can't aplly linear approximation whenx0=1 for square root(28), it should be that x0=5 or x0=6

Answer 3

it's still illegible question