linear approximation

[nat]

Let f(x)=x^3. The equation of the tangent line to f(x) at x=5 is y= __?


Using this, we find our approximation for 4.7^3 is ___


not sure how to work this out...

 

[Deleted member 4993]

Let f(x)=x^3. The equation of the tangent line to f(x) at x=5 is y= __?


Using this, we find our approximation for 4.7^3 is ___


not sure how to work this out...

To find the equation of a line - you need to know the slope and a point on the line.

How are the slope of a tangent line (of a curve) and the derivative of a function (representing the curve) related to each other?

What is the co-ordinate of the tangent point (through which tangent point must pass)?

Use Taylor's series approximation for finding the value of 4.7^3.

 

[soroban]

Hello, nat!

\(\displaystyle \text{Let }f(x)\:=\:x^3\)
\(\displaystyle \text{Find the equation of the tangent line to }f(x)\text{ at }x=5.\)

\(\displaystyle \text{First: }\,f(5) \,=\,5^3 \,=\,25.\)
\(\displaystyle \text{We have the point of tangency: }(5,\,125)\)

\(\displaystyle \text{The derivative is: }\,f'(x) \,=\,3x^2\)
\(\displaystyle \text{Then: }\.f'(5) \,=\,3(5^2) \,=\,75\)
\(\displaystyle \text{We have the slope of the tangent: }\,m \,=\,75\)

\(\displaystyle \text{The equation of the tangent is: }\:y - 125 \:=\:75(x - 5) \quad\Rightarrow\quad y \:=\:75x - 250\)

\(\displaystyle \text{Using this, find an approximation for }4.7^3\)

\(\displaystyle \text{In the equation of the tangent, let }x = 4.7\)

\(\displaystyle \text{We have: }\,y \:=\:75(4.7) - 250 \:=\:102.5\)

\(\displaystyle \text{Therefore: }\:4.7^3 \;\approx\;102.5\)


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This is the idea behind this approximation.


\(\displaystyle \text{We have the graph of the function: }\,f(x) \,=\,x^3\)

\(\displaystyle \text{We can find the value of }\,f(5) \,=\,5^3 \,=\,125\)

\(\displaystyle \text{We want an approximate value of }f(4.7) \,=\,4.7^3\)
. . \(\displaystyle without\text{ doing any cubing.}\)

\(\displaystyle \text{We can graph }f(x) \,=\,x^3\text{ and locate }P(5,\,125).\)

        |
        |                o
        |
        |              o
        |           P o
        |       Q   o
        |       o   :
        o       :   :
        |      ?:   :125
        |       :   :
        |       :   :
    . - + - - - + - + - - - - - - -
        |      4.7  5

\(\displaystyle \text{And we want the height at }Q\text{ where }x = 4.7\)



\(\displaystyle \text{Draw the tangent at }P(5,125)\)

        |
        |                o  *
        |                 *
        |              o*
        |          P o*
        |       Q   o
        |       o * :
        o       *   :
        |     * :   :
        |   *   :   :
        | *     :   :
    - - + - - - + - + - - - - - - -
        |      4.7  5

\(\displaystyle \text{Instead of finding the exact height of }Q,\)
. . \(\displaystyle \text{we use the height of }x\text{ on the tangent line.}\)

        |
        |                o  *
        |                 *
        |              o*
        |          P o*
        |       Q   o
        |       o * :
        o       x   :
        |     * :   :
        |   *   :   :
        | *     :   :
    - - + - - - + - + - - - - - - -
        |      4.7  5

\(\displaystyle \text{And this serves as a fair approximation of the height at }Q.\)

 

[Deleted member 4993]

Soroban is exactly right - knowledge of Taylor's series is not needed. Taylor's series can be used - but not needed.

 

[nat]

THANK YOU!