Question

1. $f(x) = 2x^3 - 7x^2 + 6$

domain: $(-infty, infty)$
range: $(-infty, infty)$
rel. maximum(s):
rel. minimum(s):
end behavior: as $x \to -infty$, $f(x) \to -infty$
as $x \to infty$, $f(x) \to infty$
inc. intervals:
dec. intervals:

Explanation:

Step1: Find the derivative of the function

The function is \(f(x)=2x^{3}-7x^{2}+6\). Using the power rule \((x^{n})^\prime = nx^{n - 1}\), the derivative \(f^\prime(x)=6x^{2}-14x=2x(3x - 7)\).

Step2: Find the critical points

Set \(f^\prime(x)=0\). So \(2x(3x - 7)=0\). Solving \(2x=0\) gives \(x = 0\), and solving \(3x-7=0\) gives \(x=\frac{7}{3}\).

Step3: Use the first - derivative test

• Test the interval \((-\infty,0)\):

Let \(x=-1\), then \(f^\prime(-1)=6\times(-1)^{2}-14\times(-1)=6 + 14=20>0\). So the function is increasing on \((-\infty,0)\).

• Test the interval \((0,\frac{7}{3})\):

Let \(x = 1\), then \(f^\prime(1)=6\times1^{2}-14\times1=6-14=-8<0\). So the function is decreasing on \((0,\frac{7}{3})\).

• Test the interval \((\frac{7}{3},\infty)\):

Let \(x = 3\), then \(f^\prime(3)=6\times3^{2}-14\times3=54 - 42 = 12>0\). So the function is increasing on \((\frac{7}{3},\infty)\).

Since the function changes from increasing to decreasing at \(x = 0\), the relative maximum is \(f(0)=2\times0^{3}-7\times0^{2}+6=6\).
Since the function changes from decreasing to increasing at \(x=\frac{7}{3}\), the relative minimum is \(f(\frac{7}{3})=2\times(\frac{7}{3})^{3}-7\times(\frac{7}{3})^{2}+6=2\times\frac{343}{27}-7\times\frac{49}{9}+6=\frac{686}{27}-\frac{343}{9}+6=\frac{686 - 1029+162}{27}=\frac{-181}{27}\approx - 6.7\)

Answer:

Rel. Maximum(s): \((0,6)\)
Rel. Minimum(s): \((\frac{7}{3},-\frac{181}{27})\)
Inc. Intervals: \((-\infty,0)\cup(\frac{7}{3},\infty)\)
Dec. Intervals: \((0,\frac{7}{3})\)