The linear approximation at x = 0 to f(x) = \sqrt { 5 + 4 x } is y =
at x=0 the slope is 2/√5
at x=0, y = √5
So, the line there is y = (2/√5)x + √5
http://www.wolframalpha.com/input/?i=plot+y%3D%E2%88%9A%285%2B4x%29%2Cy%3D2%2F%E2%88%9A5+x%2B%E2%88%9A5+for+x+%3D+-2+to+2
To find the linear approximation at x = 0 to the function f(x) = √(5 + 4x), we can use the concept of the tangent line.
The linear approximation is given by the equation of the tangent line to the graph of the function at the point (x, f(x)) = (0, √5). The equation of a line can be written in the form y = mx + b, where m is the slope of the line and b is the y-intercept.
First, let's find the slope of the tangent line. The slope of the tangent line at a point (x0, f(x0)) on a curve can be found using the derivative of the function at that point.
The derivative of the function f(x) = √(5 + 4x) can be found using the power rule for differentiation. Applying the power rule, we have:
f'(x) = (1/2)(5 + 4x)^(-1/2)(4) = 2/(√(5 + 4x))
Evaluating the derivative at x = 0, we have:
f'(0) = 2/(√5)
Therefore, the slope of the tangent line at x = 0 is m = 2/(√5).
Next, let's find the y-intercept of the tangent line. We already have a point on the line, (x, f(x)) = (0, √5). We can substitute these values into the equation y = mx + b and solve for b:
√5 = (2/(√5))(0) + b
√5 = b
Therefore, the y-intercept of the tangent line is b = √5.
Now we can write the equation of the tangent line, or the linear approximation, in the form y = mx + b:
y = (2/(√5))x + √5
So, the linear approximation at x = 0 to the function f(x) = √(5 + 4x) is y = (2/(√5))x + √5.