# Find the linearization L(x) of the function at a. f(x) = x2/3, a = 54

## Answers

-6 es no comprendo perro la sala.

[tex]\displaystyle L(x)=1+\frac{1}{4}x[/tex]

Step-by-step explanation:

Linearization

It consists in finding a linear approximate function around certain point x=a of a non linear function. The equation used to linearize a function f is given by

[tex]L(x)=f(a)+f'(a)(x-a)[/tex]

Where f'(a) is the first derivative of f at x=a

The questions gives us this function

[tex]f(x)=\sqrt{x}[/tex]

And the point a=4. Let's compute f(a)

[tex]f(4)=\sqrt{4}=2[/tex]

We are using only the positive root. Now we compute the derivative

[tex]\displaystyle f'(x)=\frac{1}{2\sqrt{x}}[/tex]

[tex]\displaystyle f'(4)=\frac{1}{2\sqrt{4}}[/tex]

[tex]\displaystyle f'(4)=\frac{1}{4}[/tex]

Replacing the values into the equation

[tex]\displaystyle L(x)=2+\frac{1}{4}(x-4)[/tex]

[tex]\displaystyle L(x)=1+\frac{1}{4}x[/tex]

Step-by-step explanation:

The equation of the tangent line at x=1 can be written in point-slope form as

... L(x) = f'(1)(x -1) +f(1)

The derivative is ...

... f'(x) = 4x^3 +4x

so the slope of the tangent line is f'(1) = 4+4 = 8.

The value of the function at x=1 is

... f(1) = 1^4 +2·1^2 = 3

So, your linearization is ...

... L(x) = 8(x -1) +3

or

... L(x) = 8x -5

[tex]Find the linearization l(x) of the function at a. f(x) = x^4 + 2x^2, a = 1[/tex]

The formula of linearization is

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

We have [tex]f(x) = x^4+4x^2[/tex]

And it is given in the question that a=1.

So f(a) = f(1) = 1+4=5

Now we need to find f'(a) and for that , first we need to find f'(x).

[tex]f'(x) = 4x^{4-1}+4(2)x^{2-1}[/tex]

[tex]f'(x)=4x^3+8x[/tex]

At a=1,

[tex]f'(1)= 4+8=12[/tex]

Substituting the values of f(1) and f'(1) in the linearization formula

L=5+12(x-1)

L= 5+12x-12

L= 12x -7

And that's the required linearization .

As is it tangent line

by using slope formula

y-y1 = m(x-x1)

given a=pi/6=x1

f(x)= sinx

above equation can be wriiten as

f(x)-f(a) = f’(a)(x – a)

so L(x)=f’(a) (x-a) + f(a)

f(x) = sinx =f(pi/6) = 1/2= y1

f’(x) = cosx f’(pi/6)= 3/2 = m

now putting values

L(x) = √ 3/2( x-pi/6)+1/2

L(x) = (√ 3/2)x+6-pi√ 3/12

Please answer please thank you so you

Please note that your x^3/4 is ambiguous. Did you mean (x^3) divided by 4

or did you mean x to the power (3/4)? I will assume you meant the first, not the second. Please use the "^" symbol to denote exponentiation.

If we have a function f(x) and its derivative f'(x), and a particular x value (c) at which to begin, then the linearization of the function f(x) is

f(x) approx. equal to [f '(c)]x + f(c)].

Here a = c = 81.

Thus, the linearization of the given function at a = c = 81 is

f(x) (approx. equal to) 3(81^2)/4 + [81^3]/4

Note that f '(c) is the slope of the line and is equal to (3/4)(81^2), and f(c) is the function value at x=c, or (81^3)/4.

What is the linearization of f(x) = (x^3)/4, if c = a = 81?

It will be f(x) (approx. equal to)

The equation is a vertical line, so the slope is infinite.

m= infinite, b=x * 4 + 2x * 2