What is the equation of the tangent line of f(x)=3x^2 at x=1?

Mar 29, 2018

$y = 6 x - 3$

Explanation:

The first objective is to find the slope of the tangent line to the curve of $f \left(x\right) = 3 {x}^{2}$ at $x = 1.$ This can be done by finding $f ' \left(1\right)$, as the derivative at a point represents the slope of the tangent line to the curve at that point (graphically interpreted).

$f ' \left(x\right) = \left(3\right) \left(2\right) {x}^{2 - 1}$

$f ' \left(x\right) = 6 x$

$f ' \left(1\right) = 6$ is the slope of the tangent line at $x = 1.$

Now we use the point-slope form of a line to find the tangent line equation:

$y - {y}_{0} = m \left(x - {x}_{0}\right)$ where $\left({x}_{0} , {y}_{0}\right)$ is a point on the line and $m$ is the slope.

We're given ${x}_{0} = 1 ,$ so ${y}_{0} = f \left(1\right) = 3 \left({1}^{2}\right) = 3$

$m = 6 ,$ as calculated above.

$y - 3 = 6 \left(x - 1\right)$

$y - 3 = 6 x - 6$

$y = 6 x - 3$

Mar 29, 2018

$y = 6 x - 3$

Explanation:

We know that the function and it's tangent have the same gradient, so at $f \left(1\right)$ both will have the same gradient

To get the gradient function of a function we know it is merely the first derivative

$f ' \left(x\right) = 6 x$

So now we want to find on our gradient function what the actual gradient will be at one

$f ' \left(1\right) = 6 \cdot 1 = 6$

So as the tangent is a straight line the formula for it is $y = m x + c$ with m being the gradient

So as the two gradients are equal $6 = m$

$y = 6 x + c$

Now all we have to find out is the y-intercept to do this we have to substiute a co-ordinate from the tangent in. We know that the tangent touches the function when $x = 1$

So we can use that co-ordinate pair from the original function

$f \left(1\right) = 3 {\left(1\right)}^{2} = 3$
$\left(1 , 3\right)$

Substitue that into the tangent equation to fine c

$3 = 6 \left(1\right) + c$ so $- 3 = c$

$y = 6 x - 3$