# equation of tangent line

• April 27th 2010, 11:41 PM
serious331
equation of tangent line
define $f(x) = x^3+2x+1$ for all x. Find the equation of the tangent line to the graph of $f:R \rightarrow R$ at the point (2,13).

I know that for a function $f:I \rightarrow R$ (I is a neighborhood of the point $x_0$) that is differentiable at $x_0$, the tangent to the graph of the point $({x_0},(f(x_0))$ is given by the equation:

$y = f(x_0)+f'(x_0)(x-{x_0})$

To find the slope of the line, wehave

$\lim_{x->{x_0}} \frac{f(x)-f(x_0)}{x-{x_0}}$

$= \lim_{x->{x_0}} \frac{(x^3+2x+1)-({x_0}^3+2{x_0}+1)}{x-{x_0}}$

$= \lim_{x->{x_0}} (x^3-{x_0}^3)+2 = 2$

is the slope of this function is 2 then? I cannot figure out if this is correct and how to find the equation of the tangent line (Headbang)
• April 28th 2010, 12:20 AM
harish21
Quote:

Originally Posted by serious331
define $f(x) = x^3+2x+1$ for all x. Find the equation of the tangent line to the graph of $f:R \rightarrow R$ at the point (2,13).

I know that for a function $f:I \rightarrow R$ (I is a neighborhood of the point $x_0$) that is differentiable at $x_0$, the tangent to the graph of the point $({x_0},(f(x_0))$ is given by the equation:

$y = f(x_0)+f'(x_0)(x-{x_0})$

To find the slope of the line, wehave

$\lim_{x->{x_0}} \frac{f(x)-f(x_0)}{x-{x_0}}$

$= \lim_{x->{x_0}} \frac{(x^3+2x+1)-({x_0}^3+2{x_0}+1)}{x-{x_0}}$

$= \lim_{x->{x_0}} (x^3-{x_0}^3)+2 = 2$

is the slope of this function is 2 then? I cannot figure out if this is correct and how to find the equation of the tangent line (Headbang)

The slope of your line is found as:

$\lim_{x->2} \frac{f(x)-f(2)}{x-2} = \lim_{x->2} \frac{(x^3+2x+1)-(8+4+1)}{x-2} = \lim_{x->2} \frac{x^3+2x-12}{x-2} =14$

so $f'(x_0)=14$

now find the equation of the tangent using the formula you have!
• April 28th 2010, 05:16 AM
HallsofIvy
Quote:

Originally Posted by serious331
define $f(x) = x^3+2x+1$ for all x. Find the equation of the tangent line to the graph of $f:R \rightarrow R$ at the point (2,13).

I know that for a function $f:I \rightarrow R$ (I is a neighborhood of the point $x_0$) that is differentiable at $x_0$, the tangent to the graph of the point $({x_0},(f(x_0))$ is given by the equation:

$y = f(x_0)+f'(x_0)(x-{x_0})$

To find the slope of the line, wehave

$\lim_{x->{x_0}} \frac{f(x)-f(x_0)}{x-{x_0}}$

$= \lim_{x->{x_0}} \frac{(x^3+2x+1)-({x_0}^3+2{x_0}+1)}{x-{x_0}}$

$\frac{(x^2- x_0^3)+ 2(x- x_0)}{x- x_0}= \frac{(x- x_0)(x^2+ xx_0+ x_0^2)- 2(x- x_0)}{x- x_0}= x^2+ xx_0+ x_0^2+ 2$ and, as x goes to $x_0$, that becomes $3x_0+ 2$. Although, as harish21 says, it is simpler to put the [tex]x_0= 2[tex], $f(x_0)= 13$ from the start.

Quote:

$= \lim_{x->{x_0}} (x^3-{x_0}^3)+2 = 2$

is the slope of this function is 2 then? I cannot figure out if this is correct and how to find the equation of the tangent line (Headbang)
You appear to be saying that $\frac{(x^3- x_0^2)+ 2(x- x_0)}{x- x_0}= x^3- x_0^3+ 2$. That is, that you are canceling the " $x- x_0$ only from second term. You can't do that, of course, $\frac{a+ b}{b}$ is NOT equal to a+ 1.

Once you have that, use the formula you gave: $y= f(x_0)+ f'(x_0)(x- x_0)$ with $x_0= 2$, $f(x_0)= 13$, and $f'(x_0)= 14$.