Rocket's path is graphed by function $f(x) = x^3 - 8x$

Question

A rocket's path is graphed by the function $f(x) = x^3 - 8x$, where $x$ refers to time after launch (s). At what point after its launch would you release its thrusters in order for its tangent path to pass through the point $(4,0)?$

Answer 1

Better to use different symbols $x$ for distance and $t$ for time

If the rocket is headed only to look at the point $(4,0)$ ( reaching there is another matter due to a different trajectory )..

match the slope of cubic's derivative to slope at the required point

$$\dfrac{(x^3-8x)-0}{x-4}= 3 x^2-8$$

simplify /solve cubic (edit: with simplification error corrected )

$$(x-2)(x^2+8)=0, \rightarrow x =2 $$

is the time for it to look at $(4,0),$ and fire the rocket imparting a very high velocity like a laser ray.

Answer 2

Assuming that the position is $(x,f(x))$

• consider a generic point $P(x,f(x))$ on the path
• set the condition for which $f'(x)$ is equal to the slope $PQ$ with $Q=(4,0)$

that is

$$\dfrac{(x^3-8x)-0}{x-4}= 3 x^2-8 \iff x^3-8x=3x^3-8x-12x^2+32 \iff2x^3-12x^2+32=0$$

from which we obtain $x=2$.