Question

find the absolute extrema of the function (if any exist) on each interval. (if an answer does not exist, enter dne.)
$f(x)=x^{2}-8x$
(a) $-1,8$
minimum $(x,y)=(\boxed{4,-16})$
maximum $(x,y)=(\boxed{-1,9})$
(b) $(4,9$
minimum $(x,y)=(\boxed{dne})$
maximum $(x,y)=(\boxed{9,9})$
(c) $(0,8)$
minimum $(x,y)=(\boxed{dne})$
maximum $(x,y)=(\boxed{dne})$
(d) $4,10)$
minimum $(x,y)=(\boxed{4,-16})$
maximum $(x,y)=(\boxed{dne})$

Explanation:

Step1: Find the derivative of the function

The function is \(f(x)=x^{2}-8x\). Using the power rule \((x^{n})^\prime = nx^{n - 1}\), the derivative \(f^\prime(x)=2x - 8\).

Step2: Find the critical points

Set \(f^\prime(x)=0\), so \(2x-8 = 0\). Solving for \(x\):

$$
LATEXBLOCK0
$$

Step3: Analyze the interval \((0,8)\)

• The critical point \(x = 4\) is in the open interval \((0,8)\). But for an absolute minimum or maximum in an open interval \((a,b)\), the function values at the endpoints \(a\) and \(b\) are not included (since the interval does not contain \(0\) and \(8\)).
• We check the behavior of the function. The function \(y=x^{2}-8x=(x - 4)^{2}-16\) is a parabola opening upwards (because the coefficient of \(x^{2}\) is positive, \(a = 1>0\)).
• As \(x\to0^{+}\), \(y=x(x - 8)\to0\). As \(x\to8^{-}\), \(y=x(x - 8)\to0\). And \(f(4)=(4)^{2}-8\times4=16-32=-16\). But since \(x = 4\) is the only critical point in the interval \((0,8)\) and the interval is open, there is no absolute minimum (because we can get arbitrarily close to \(x = 4\) from the right and left, but the point \(x = 4\) is not the endpoint of the open - interval \((0,8)\)) and no absolute maximum (since we can approach the values near \(x=0\) and \(x = 8\) but not reach them).

Answer:

For part (c) \((0,8)\):

• minimum \((x,y)=\text{DNE}\)
• maximum \((x,y)=\text{DNE}\)