Tangent line of $f(x)=x^3-x+5$ at $x=2$

Question

Find the tangent line of the function : $$f(x)=x^3-x+5$$ at the point $x=2$.

Having difficult time solving this :( Does anyone know the way to solve this sort of problem?? Wish somebody could help me

Answer 1

HINT (if you're familiar with elementary differentation) :

The tangent line at a point $A=(x_0,f(x_0))$ of a differentiable function $f$ at $x=x_0$, is given by :

$$y-f(x_0)=f'(x_0)(x-x_0)$$

Obviously, the function $f(x) = x^3-x + 5$ is differentiable at $x=2$. Can you now use the formula above to find the tangent line ? What term expresses the slope of it ?

Alternative (without derivatives) :

If $f$ is defined on an open interval that contains $c$ and if the limit $$\lim_{Δx \to 0} \frac{Δy}{Δx} = \lim_{Δx\to 0} \frac{f(c+Δx) - f(c)}{Δx}=m \space \in \space \mathbb R $$ then the line passing through $(c,f(c))$ with slope $m$ is the tangent line to the graph of $f$ at the point $(c,f(c))$.

Answer 2

To find the tangent, we need the gradient (rate of change) at $x=2$.

\begin{align} f(x) & = x^3 - x + 5 \\ f'(x) & = 3x^2 - 1 \\ f'(2) & = 3(2)^2 - 1 \\ & = 11 \end{align}

Hence the gradient of the tangent is $11$. But, we need the y-coords of $f(x)$ at $x=2$.

\begin{align} f(x) & = x^3 - x + 5 \\ f(2) & = 2^3 - 2 + 5 \\ & = 11 \\ \end{align}

\begin{align} \text{Since}~& t(x) ~\text{passes through (2,11)}\\ t(x) & = mx+c \\ 11 & = 11(2)+c \\ c&=-11 \\ \end{align}

Hence the equation of the tangent at $x=2$ is $t(x)=11x-11$.