Chapter 4.4, Problem 1QR

To determine

The local extreme values of the given function using the first derivative test.

Answer to Problem 1QR

It has been determined that the given function has no local extreme values.

Explanation of Solution

Given:

The function, $y=x^3-6x^2+12x-8$ .

Concept used:

The critical points of a function $y\left(x\right)$ are the points $x=c$ such that $y^{\prime }\left(c\right)=0$ .

If $y^{\prime }\left(x\right)<0$ for $x<c$ and $y^{\prime }\left(x\right)>0$ for $x>c$ , then the function $y\left(x\right)$ has a local minima at $x=c$ and vice versa.

Calculation:

The given function is $y=x^3-6x^2+12x-8$ .

Differentiating,

  $y^{\prime }\left(x\right)=3x^2-12x+12$

Simplifying,

  $y^{\prime }\left(x\right)=3\left(x^2-4x+4\right)$

On further simplification,

  $y^{\prime }\left(x\right)=3{\left(x-2\right)}^2$

Equating the first derivative to zero to obtain the critical points,

  $3{\left(x-2\right)}^2=0$

Solving, $x=2$ .

So, the only critical point of the given function is $x=2$ .

Note that, $y^{\prime }\left(x\right)=3{\left(x-2\right)}^2>0$ for $x<2$ and for $x>2$ .

This implies that there is no point of local extrema at $x=2$ .

Since there are no other critical points, it follows that the given function has no local extreme values.

Conclusion:

It has been determined that the given function has no local extreme values.