Find the linearization L(x) of the function at a.

f(x) = x^1/2, a = 25

Well, well, well, let's find the linearization of this square root function at the point x = 25. Now, in order to do that, we need to find the derivative of the function first.

The derivative of f(x) = x^(1/2) can be found using the power rule for differentiation. So, buckle up and put on your math cap, because here we go!

f'(x) = (1/2) * x^(-1/2)

Great! Now, let's substitute a in place of x to find the slope of the tangent line at a.

f'(a) = (1/2) * a^(-1/2)
= (1/2) * 25^(-1/2)
= (1/2) * 1/5
= 1/10

Now, we can plug this slope and the point (25, f(25)) = (25, 5) into the point-slope form of a linear equation to find the linearization L(x):

L(x) = f(a) + f'(a) * (x - a)
= 5 + (1/10) * (x - 25)

Ta-da! There's your linearization, my friend. L(x) = 5 + (1/10) * (x - 25).

But hey, don't forget to take these calculations with a grain of salt. After all, life is full of uncertainties, and math is no exception!

To find the linearization L(x) of the function f(x) = x^(1/2) at a = 25, we can use the formula:

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

First, let's find f(a) and f'(a):

f(a) = a^(1/2) = 25^(1/2) = 5

To find f'(a), we'll need to differentiate f(x) = x^(1/2) using the power rule. The power rule states that if f(x) = x^n, then f'(x) = nx^(n-1).

Here, n = 1/2:
f'(x) = (1/2)x^((-1/2) - 1)
f'(x) = (1/2)x^(-3/2)

Now, let's find f'(a):
f'(a) = (1/2)a^(-3/2) = (1/2)(25)^(-3/2) = (1/2) * (1/25^(3/2)) = 1/50

Now we can substitute the values of f(a), f'(a), and a into the linearization formula:

L(x) = f(a) + f'(a)(x - a)
L(x) = 5 + (1/50)(x - 25)
L(x) = 5 + (1/50)x - (1/50) * 25
L(x) = 5 + (1/50)x - 1/2
L(x) = (1/50)x + 9/2

Therefore, the linearization L(x) of the function f(x) = x^(1/2) at a = 25 is L(x) = (1/50)x + 9/2.

To find the linearization L(x) of the function f(x) at a, we need to determine the equation of the tangent line to the graph of f(x) at the point (a, f(a)).

The equation of a line in point-slope form is given by:
y - y1 = m(x - x1)

Where (x1, y1) represents a point on the line and m is the slope of the line.

In this case, the point (x1, y1) is (a, f(a)) = (25, √25) = (25, 5).

To find the slope of the tangent line, we can use the derivative of the function f(x).

Taking the derivative of f(x) = x^(1/2) with respect to x:

f'(x) = (1/2)x^(-1/2)

Evaluating f'(x) at x = a:

f'(a) = (1/2)(25)^(-1/2) = (1/2)(1/5) = 1/10

Therefore, the slope of the tangent line is m = 1/10.

Now, we can use the point-slope form to find the equation of the tangent line:

y - 5 = (1/10)(x - 25)

Simplifying the equation:

y - 5 = (1/10)x - 5/2

y = (1/10)x + 5 - 5/2

y = (1/10)x + 15/2

So, the linearization L(x) of the function f(x) at a = 25 is given by the equation:

L(x) = (1/10)x + 15/2

What you want is a line that approximates the curve at some point (and, by extension, in some small interval).

f'(x) = 1/(2√x)
f'(25) = 1/10

So, you want the line through (25,5) with slope = 1/10

(y-5) = 1/10 (x-25)
y = x/10 + 5/2

If you restrict your interval of x-values, you can show that the line is as close as you want to the curve in that interval.