3. sketch the graph of the following function $f(x)=2x^{3}-9x^{2}+12x - 3$ by completing the following…

Explanation:

Step1: Analyze asymptotes and end - behavior

The function $f(x)=2x^{3}-9x^{2}+12x - 3$ is a polynomial function of degree 3. Polynomial functions have no vertical or horizontal asymptotes. As $x\to+\infty$, $y = 2x^{3}-9x^{2}+12x - 3\to+\infty$ since the leading term $2x^{3}$ dominates and has a positive coefficient. As $x\to-\infty$, $y=2x^{3}-9x^{2}+12x - 3\to-\infty$.

Step2: Find the first - derivative

Differentiate $f(x)$ using the power rule. $f'(x)=6x^{2}-18x + 12=6(x^{2}-3x + 2)=6(x - 1)(x - 2)$.

Step3: Determine increasing and decreasing intervals and local extrema

Set $f'(x)=0$. Then $6(x - 1)(x - 2)=0$, so $x = 1$ and $x = 2$. Test intervals:

• For $x\lt1$, let $x = 0$. Then $f'(0)=6(0 - 1)(0 - 2)=12\gt0$, so $f(x)$ is increasing on $(-\infty,1)$.
• For $1\lt x\lt2$, let $x=\frac{3}{2}$. Then $f'(\frac{3}{2})=6(\frac{3}{2}-1)(\frac{3}{2}-2)=6\times\frac{1}{2}\times(-\frac{1}{2})=- \frac{3}{2}\lt0$, so $f(x)$ is decreasing on $(1,2)$.
• For $x\gt2$, let $x = 3$. Then $f'(3)=6(3 - 1)(3 - 2)=12\gt0$, so $f(x)$ is increasing on $(2,+\infty)$. Local maximum at $x = 1$: $f(1)=2(1)^{3}-9(1)^{2}+12(1)-3=2 - 9 + 12-3=2$. Local minimum at $x = 2$: $f(2)=2(2)^{3}-9(2)^{2}+12(2)-3=16-36 + 24-3=1$.

Step4: Find the second - derivative

Differentiate $f'(x)$: $f''(x)=12x-18 = 6(2x - 3)$.

Step5: Determine concavity and inflection points

Set $f''(x)=0$. Then $6(2x - 3)=0$, so $x=\frac{3}{2}$. Test intervals:

• For $x\lt\frac{3}{2}$, let $x = 1$. Then $f''(1)=12\times1-18=-6\lt0$, so $f(x)$ is concave down on $(-\infty,\frac{3}{2})$.
• For $x\gt\frac{3}{2}$, let $x = 2$. Then $f''(2)=12\times2-18 = 6\gt0$, so $f(x)$ is concave up on $(\frac{3}{2},+\infty)$. The inflection point is at $x=\frac{3}{2}$, and $f(\frac{3}{2})=2(\frac{3}{2})^{3}-9(\frac{3}{2})^{2}+12(\frac{3}{2})-3=\frac{27}{4}-\frac{81}{4}+18 - 3=\frac{27 - 81+72 - 12}{4}=\frac{6}{4}=\frac{3}{2}$.

Step6: Sketch the graph

Plot the local maximum $(1,2)$, local minimum $(2,1)$, inflection point $(\frac{3}{2},\frac{3}{2})$. Use the information about increasing/decreasing intervals and concavity to sketch the curve.

Answer:

• Asymptotes: None. End - behavior: As $x\to+\infty$, $y\to+\infty$; as $x\to-\infty$, $y\to-\infty$.
• Increasing intervals: $(-\infty,1)$ and $(2,+\infty)$. Decreasing interval: $(1,2)$. Local maximum: $(1,2)$. Local minimum: $(2,1)$.
• Concave up interval: $(\frac{3}{2},+\infty)$. Concave down interval: $(-\infty,\frac{3}{2})$. Inflection point: $(\frac{3}{2},\frac{3}{2})$.
• The graph is sketched using the above - mentioned points and intervals.