Answer:
[tex]\textsf{(a)} \quad \textsf{Increasing}: \quad \left(-\infty, \dfrac{-1-\sqrt{13}}{2}\right) \cup \left(\dfrac{-1+\sqrt{13}}{2}, \infty\right)[/tex]
[tex]\textsf{Decreasing}: \quad \left(\dfrac{-1-\sqrt{13}}{2} < x < \dfrac{-1+\sqrt{13}}{2}\right)[/tex]
[tex]\textsf{(b)} \quad \textsf{Minimum}: \quad \left(\dfrac{-1-\sqrt{13}}{2},\dfrac{19+13\sqrt{13}}{2}\right)[/tex]
[tex]\textsf{Maximum}: \quad \left(\dfrac{-1+\sqrt{13}}{2},\dfrac{19-13\sqrt{13}}{2}\right)[/tex]
[tex]\textsf{(c)} \quad \textsf{Point of inflection}: \quad \left(-\dfrac{1}{2}, \dfrac{19}{2}\right)[/tex]
[tex]\textsf{Concave up}: \quad \left(-\dfrac{1}{2}, \infty\right)[/tex]
[tex]\textsf{Concave down}: \quad \left(- \infty,-\dfrac{1}{2}\right)[/tex]
Step-by-step explanation:
Given function:
[tex]f(x) = 2x^3 + 3x^2-18x[/tex]
[tex]\textsf{A function is \textbf{increasing} when the \underline{gradient is positive}}\implies f'(x) > 0[/tex]
[tex]\textsf{A function is \textbf{decreasing} when the \underline{gradient is negative}} \implies f'(x) < 0[/tex]
Differentiate the given function:
[tex]\implies f'(x)=3 \cdot 2x^{3-1}+2 \cdot 3x^{2-1}-18x^{1-1}[/tex]
[tex]\implies f'(x)=6x^2+6x-18[/tex]
Complete the square:
[tex]\implies f'(x)=6(x^2+x-3)[/tex]
[tex]\implies f'(x)=6\left(x^2+x+\dfrac{1}{4}-3-\dfrac{1}{4}\right)[/tex]
[tex]\implies f'(x)=6\left(x^2+x+\dfrac{1}{4}\right)-\dfrac{39}{2}[/tex]
[tex]\implies f'(x)=6\left(x+\dfrac{1}{2}\right)^2-\dfrac{39}{2}[/tex]
Increasing
To find the interval where f(x) is increasing, set the differentiated function to more than zero:
[tex]\implies f'(x) > 0[/tex]
[tex]\implies 6\left(x+\dfrac{1}{2}\right)^2-\dfrac{39}{2} > 0[/tex]
[tex]\implies 6\left(x+\dfrac{1}{2}\right)^2 > \dfrac{39}{2}[/tex]
[tex]\implies \left(x+\dfrac{1}{2}\right)^2 > \dfrac{39}{12}[/tex]
[tex]\implies \left(x+\dfrac{1}{2}\right)^2 > \dfrac{13}{4}[/tex]
[tex]\textsf{For\;\;$u^n > a$,\;\;if\;$n$\;is\;even\;then\;\;$u < -\sqrt[n]{a}$\;\;or\;\;$u > \sqrt[n]{a}$}.[/tex]
Therefore:
[tex]x+\dfrac{1}{2} < -\sqrt{\dfrac{13}{4}}\implies x < \dfrac{-1-\sqrt{13}}{2}[/tex]
[tex]x+\dfrac{1}{2} > \sqrt{\dfrac{13}{4}}\implies x < \dfrac{-1+\sqrt{13}}{2}[/tex]
So the interval on which function f(x) is increasing is:
[tex]\left(-\infty, \dfrac{-1-\sqrt{13}}{2}\right) \cup \left(\dfrac{-1+\sqrt{13}}{2}, \infty\right)[/tex]
Decreasing
To find the interval where f(x) is decreasing, set the differentiated function to less than zero:
[tex]\implies f'(x) < 0[/tex]
[tex]\implies 6\left(x+\dfrac{1}{2}\right)^2-\dfrac{39}{2} < 0[/tex]
[tex]\implies \left(x+\dfrac{1}{2}\right)^2 < \dfrac{13}{4}[/tex]
[tex]\textsf{For\;\;$u^n < a$,\;\;if\;$n$\;is\;even\;then\;\;$-\sqrt[n]{a} < u < \sqrt[n]{a}$}.[/tex]
Therefore:
[tex]x+\dfrac{1}{2} > -\sqrt{\dfrac{13}{4}}\implies x > \dfrac{-1-\sqrt{13}}{2}[/tex]
[tex]x+\dfrac{1}{2} < \sqrt{\dfrac{13}{4}}\implies x < \dfrac{-1+\sqrt{13}}{2}[/tex]
So the interval on which function f(x) is decreasing is:
[tex]\left(\dfrac{-1-\sqrt{13}}{2} < x < \dfrac{-1+\sqrt{13}}{2}\right)[/tex]
To find x-coordinates of the local minimum and maximum set the differentiated function to zero and solve for x:
[tex]\implies f'(x)=0[/tex]
[tex]\implies 6\left(x+\dfrac{1}{2}\right)^2-\dfrac{39}{2} =0[/tex]
[tex]\implies \left(x+\dfrac{1}{2}\right)^2 =\dfrac{13}{4}[/tex]
[tex]\implies x=\dfrac{-1-\sqrt{13}}{2},\;\;\dfrac{-1+\sqrt{13}}{2}[/tex]
To find the y-coordinates of the turning points, substitute the found values of x into the function and solve for y:
[tex]\implies f\left(\dfrac{-1-\sqrt{13}}{2}\right)=\dfrac{19+13\sqrt{13}}{2}[/tex]
[tex]\implies f\left(\dfrac{-1+\sqrt{13}}{2}\right)=\dfrac{19-13\sqrt{13}}{2}[/tex]
Therefore:
[tex]\textsf{Minimum}: \quad \left(\dfrac{-1-\sqrt{13}}{2},\dfrac{19+13\sqrt{13}}{2}\right) \approx (-2.30, 32.94)[/tex]
[tex]\textsf{Maximum}: \quad \left(\dfrac{-1+\sqrt{13}}{2},\dfrac{19-13\sqrt{13}}{2}\right) \approx (1.30, -13.94)[/tex]
At a point of inflection, f''(x) = 0.
To find the point of inflection, differentiate the function again:
[tex]\implies f''(x)=12x+6[/tex]
Set the second derivative to zero and solve for x:
[tex]\implies f''(x)=0[/tex]
[tex]\implies 12x+6=0[/tex]
[tex]\implies x=-\dfrac{1}{2}[/tex]
Substitute the found value of x into the original function to the find the y-coordinate of the point of inflection:
[tex]\implies f\left(-\dfrac{1}{2}\right)=\dfrac{19}{2}[/tex]
Therefore, the inflection point is:
[tex]\left(-\dfrac{1}{2}, \dfrac{19}{2}\right)[/tex]
A curve y = f(x) is concave up if f''(x) > 0 for all values of x.
A curve y = f(x) is concave down if f''(x) < 0 for all values of x.
Concave up
[tex]\implies f''(x) > 0[/tex]
[tex]\implies 12x+6 > 0[/tex]
[tex]\implies x > -\dfrac{1}{2}[/tex]
[tex]\implies \left(-\dfrac{1}{2}, \infty\right)[/tex]
Concave down
[tex]\implies f''(x) < 0[/tex]
[tex]\implies 12x+6 > 0[/tex]
[tex]\implies x < -\dfrac{1}{2}[/tex]
[tex]\implies \left(- \infty,-\dfrac{1}{2}\right)[/tex]
