Question

Calculus?

Find the equation of the tangent line T to the graph of f at the given point. Use this linear approximation to complete the table.

f(x)= 6/x^2 , (2, 3/2)

Answer 1

@eliassaab can you help me please?

Answer 2

@ganeshie8 , @CGGURUMANJUNATH , @Kainui

Answer 3

Querido Vibarguen, con mucho gusto!

Our purpose in creating this table is to discover how close an appropriate linear approximation is to actual function values.

Our function is \[y=6/x ^{2.}\]

We evaluate that function for each of the given input (x-) values, one by one:
For x=1.9, f(x)=6/(1.9^2), resulting in the point (1.9,1.662).
For x = 1.99, the value of f(x) is f(1.99), or 1.5151, resulting in the point (1.99,1.5151).
And so on. Would you please calculate f(x) for x: {2, 2.01, 2.1}?

Now we have to come up with a linear approximation T(x) for f(x) = 6/(x^2) centered on x = 2.

The general form of such a linear approximation is

\[T(x)=f(x _{0})+f'(x _{0})*(x-x _{0}).\]

Answer 4

I trust you're familiar with this equation, but will review what each of the symbols mean:

T(x) is the name (label) of an algebraic approximation of the given function f(x); it's also the equation of the tangent line to the curve of f(x) at a particular x-value, which is:
\[x _{0}=\] the specific x value at which we're finding the tangent line.

Answer 5

\[f(x _{0})\] is the value of the function at this specific x value.

Answer 6

\[f'(x _{0}) \] is the value of the DERIVATIVE of the function f(x) at the given x=value.

Answer 7

So, Vibarguen, to find values to use for filling in the rest of the table, that is, filling in the tow labeled T(x), we need to find the derivative of f(x) and to specify the x-value at which the tangent line (the linear approximation) is tangent to the curve of f(x).

Given that f(x) = 6/(x^2), find f'(x).

Next, substitute\[x _{0}=2\] for x in both f(x) and f'(x).

Could you give this a try? Try writing out the equation for T(x).
Once you have T(x) done correctly, you'll just need to evaluate T(x) at each of the given x values.

Answer 8

@mathmale Hola! muchas gracias.... haber hago lo que me pediste

Answer 9

y=6/2^2 results in (2, 1.5)
y=6/2.01^2 results in (2.01, 1.49)
y=6/2.1^2 results in (2.1, 1.35)

Answer 10

is this correct?

Answer 11

the derivative of f(x) = f'(x)= -12/x^3

Answer 12

T(x)=6/x^2(2)-12/x^3(2)∗(x−x0)

Answer 13

I'm pretty sure I messed up....

Answer 14

@mathmale

Answer 15

f'(x0) = -1.5 is this right?

Answer 16

anyone?

Answer 17

Hola, Vibarguen,
Sorry for the delay. Busy housecleaning.
Yes, you're right in stating that f'(2) = -1.5 (or -3/2).

Answer 18

Your "T(x)=6/x^2(2)-12/x^3(2)∗(x−x0)" would be a lot clearer written in the following manner:
\[T(x)=f(2)+f'(2)(x-2).\]
You've correctly evaluated both f(2) and f'(2); they are 3/2 and -3/2, respectively.
Then your tangent line equation / linear approximation is:

\[T(x)=\frac{ 3 }{ 2 }-\frac{ 3 }{ 2}(x-2).\]

Answer 19

Please do whatever you need to to understand where this equation came from and what it means.

Now, back to the original problem: You have x values as follows:
x: {1.9,1.99,2,2.01, 2.1}, and need to evaluate T(x) for each one of these 5 x values.

It'd be a lot easier if you'd first simplify T(x), as follows:
\[T(x)=\frac{ 3 }{ 2 }(1-(x-2)),\]

Answer 20

...which reduces further to
\[T(x)=\frac{ 3 }{ 2 }(2-x).\]

Answer 21

Evaluate that formula for T(x) at the 5 given x values, and then write those values into the table in the appropriate columns.

Let me know when you're finished doing this. Buen suerte!!

Answer 22

In summary:
T(x) is the linear approximation (or the tangent line) to the graph of f(x) at the point (2,3/2).

T(x) will give you approximate values of the function f(x) so long as you choose only x-values that are very close to 2 (as your five are).

Answer 23

t(1.9)=0.15
T(1.99)=0.015
T(2)=0
T(2.01)=-0.015
T(2.1)=-0.15

Answer 24

Eso esta bien?