# equation of tangent line

• Apr 27th 2010, 10:41 PM
serious331
equation of tangent line
define $\displaystyle f(x) = x^3+2x+1$ for all x. Find the equation of the tangent line to the graph of $\displaystyle f:R \rightarrow R$ at the point (2,13).

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

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

To find the slope of the line, wehave

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

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

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

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)
• Apr 27th 2010, 11:20 PM
harish21
Quote:

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

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

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

To find the slope of the line, wehave

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

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

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

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:

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

so $\displaystyle f'(x_0)=14$

now find the equation of the tangent using the formula you have!
• Apr 28th 2010, 04:16 AM
HallsofIvy
Quote:

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

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

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

To find the slope of the line, wehave

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

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

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

Quote:

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

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 $\displaystyle \frac{(x^3- x_0^2)+ 2(x- x_0)}{x- x_0}= x^3- x_0^3+ 2$. That is, that you are canceling the "$\displaystyle x- x_0$ only from second term. You can't do that, of course, $\displaystyle \frac{a+ b}{b}$ is NOT equal to a+ 1.

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