Question

Find the slope of the tangent line to the graph of fat the given point.$f\left(x\right)=2x^3,at(2,16)$.

Short answer

The slope of the tangent line is$24$ .

Step-by-step solution

Step 1. Given information.

The function here given is,

$\begin{array}{l}f\left(x\right)=2x^3\\ po\mathrm{int}=(2,16)\end{array}$

Step 2. Formula used.

The tangent line to the curve$y=f\left(x\right)$at the point$P(a,f(a\left)\right)$is the line through p with slope,

$m=\underset{x\to a}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$.

provided that this limit exists.

Step 3. Finding the slope of tangent line at given point.

Let$f\left(x\right)=2x^3$,then the slope of the tangent line at$(2,16)$is,

$\begin{array}{c}m=\underset{x\to 2}{\lim }\frac{f\left(x\right)-f\left(2\right)}{x-2}\\ =\underset{x\to 2}{\lim }\frac{\left(2x^3\right)-\left(2{\left(2\right)}^3\right)}{x-2}\\ =\underset{x\to 2}{\lim }\frac{2(x^3-2^3)}{x-2}\\ =\underset{x\to 2}{\lim }\frac{2\cancel{(x-2)}(x^2+2x+4)}{\cancel{(x-2)}}\\ =\underset{x\to 2}{\lim }2(x^2+2x+4)\\ =2(2^2+2(2)+4)\\ =24\end{array}$

Therefore, the slope of the tangent line is24 .