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

f(x) = x^4/5, a = 32

Assuming you mean x^(4/5), else x^4 is huge at x=32,

f(32) = 16
f' = 4/5 x^(-1/5)
f'(32) = 4/5 * 1/2 = 2/5

So, now you have a point (32,16) and a slope (2/5). What's the line?

Sure! Let me calculate that for you. But first, let me warn you that my calculations might come with a hint of humor.

The linearization of the function f(x) = x^(4/5) at a = 32 can be found using the formula L(x) = f(a) + f'(a)(x - a). So, we need to find the values of f(32) and f'(32) first.

To calculate f(32), we can substitute x = 32 into the function f(x) = x^(4/5):
f(32) = 32^(4/5)

Now, let me do some quick calculations in my mathematical clown brain... *humming circus music*

*few seconds later*

*Tada!*

f(32) = 16

Now, we need to find f'(32), which is the derivative of f(x) = x^(4/5). To calculate f'(32), we need to use the power rule.

*few more seconds of intense thinking*

*finger snap*

f'(32) = (4/5) * 32^(4/5 - 1)

After another round of calculations...

Voila!

f'(32) = 2.8284

Now that we have f(32) = 16 and f'(32) = 2.8284, we can use the formula L(x) = f(a) + f'(a)(x - a) to find the linearization.

L(x) = 16 + 2.8284(x - 32)

And there you have it! The linearization of the function f(x) = x^(4/5) at a = 32 is L(x) = 16 + 2.8284(x - 32).

Remember, if math ever feels boring, just add a dash of humor to make it more entertaining!

To find the linearization of a function at a specific point, we need to use the concept of the tangent line. The linearization, denoted by L(x), is the equation of the tangent line to the graph of the function at the point (a, f(a)).

First, let's find the value of f(a):
f(a) = (a)^(4/5) = (32)^(4/5) = 2^4 = 16

Next, we need to find the derivative of the function f(x) with respect to x. Let's do that:

f'(x) = d/dx [x^(4/5)]
To find the derivative, we apply the power rule for differentiation. We take the exponent (4/5) and multiply it by the coefficient in front of the variable (which is 1 in this case). Then we decrease the exponent by 1.

f'(x) = (4/5)x^(4/5 - 1)
= (4/5)x^(-1/5)

Now that we have the derivative, we can find the equation of the tangent line using the point-slope form:

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

Substituting the values we have:

L(x) = 16 + [(4/5)(32)^(-1/5)](x - 32)

Now let's simplify this equation:

L(x) = 16 + (4/5)(x - 32)^(4/5)(32)^(-1/5)

Therefore, the linearization of the function f(x) = x^(4/5) at a = 32 is:

L(x) = 16 + (4/5)(x - 32)^(4/5)(32)^(-1/5)

To find the linearization, we need to find the equation of the tangent line to the function at the point (a, f(a)). The equation of a line can be expressed in the point-slope form:

y - y1 = m(x - x1)

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

First, let's find the slope of the tangent line. The slope of a function at a point can be found by taking the derivative of the function and evaluating it at that point.

The derivative of f(x) = x^(4/5) can be found using the power rule for differentiation:

f'(x) = (4/5) * x^(-1/5)

Next, evaluate the derivative at the point a = 32:

f'(32) = (4/5) * 32^(-1/5)

Simplify the expression:

f'(32) = (4/5) * 2^(-1/5) = (4/5) * (1/2^(1/5)) = (4/5) / 2^(1/5)

Now, we have the slope of the tangent line. To find the y-intercept (b), substitute the values into the equation of the line:

f(32) = 32^(4/5)

Calculate the value:

f(32) = 32^(4/5) = 32^(4/5) = 32^(4/5) = (2^5)^(4/5) = 2^4 = 16

So the point (32, 16) lies on the tangent line. Plug the values into the equation of the line:

y - 16 = (4/5) / 2^(1/5) * (x - 32)

Simplify the equation:

y - 16 = (4/5) / 2^(1/5) * x - (4/5) / 2^(1/5) * 32

y - 16 = (4/5) / 2^(1/5) * x - (4/5) / 2^(1/5) * 32

Rearrange the equation to the point-slope form:

y = (4/5) / 2^(1/5) * x - (4/5) / 2^(1/5) * 32 + 16

Simplify further:

y = (4/5) / 2^(1/5) * x - 4/5 * 2^(4/5) + 16

This is the linearization L(x) of the function f(x) at a = 32.