Question

Consider the equation below.

f(x) = 2x^3 + 3x^2 − 18x

(a) Find the interval on which f is increasing. (Enter your answer in interval notation.)


Find the interval on which f is decreasing. (Enter your answer in interval notation.)


(b) Find the local minimum and maximum values of f.

local minimum

local maximum


(c) Find the inflection point.

(x, y) =


Find the interval on which f is concave up. (Enter your answer in interval notation.)


Find the interval on which f is concave down. (Enter your answer in interval notation.)

Answer 1

`\textsf{(a)} \quad \textsf{Increasing}: \quad \left(-\infty, (-1-√(13))/(2)\right) \cup \left((-1+√(13))/(2), \infty\right)`

`\textsf{Decreasing}: \quad \left((-1-√(13))/(2) < x < (-1+√(13))/(2)\right)`

`\textsf{(b)} \quad \textsf{Minimum}: \quad \left((-1-√(13))/(2),(19+13√(13))/(2)\right)`

`\textsf{Maximum}: \quad \left((-1+√(13))/(2),(19-13√(13))/(2)\right)`

`\textsf{(c)} \quad \textsf{Point of inflection}: \quad \left(-(1)/(2), (19)/(2)\right)`

`\textsf{Concave up}: \quad \left(-(1)/(2), \infty\right)`

`\textsf{Concave down}: \quad \left(- \infty,-(1)/(2)\right)`

Explanation:

Given function:

`f(x) = 2x^3 + 3x^2-18x`

Part (a)

`\textsf{A function is \textbf{increasing} when the \underline{gradient is positive}}\implies f'(x) > 0`

`\textsf{A function is \textbf{decreasing} when the \underline{gradient is negative}} \implies f'(x) < 0`

Differentiate the given function:

`\implies f'(x)=3 \cdot 2x^(3-1)+2 \cdot 3x^(2-1)-18x^(1-1)`

`\implies f'(x)=6x^2+6x-18`

Complete the square:

`\implies f'(x)=6(x^2+x-3)`

`\implies f'(x)=6\left(x^2+x+(1)/(4)-3-(1)/(4)\right)`

`\implies f'(x)=6\left(x^2+x+(1)/(4)\right)-(39)/(2)`

`\implies f'(x)=6\left(x+(1)/(2)\right)^2-(39)/(2)`

Increasing

To find the interval where f(x) is increasing, set the differentiated function to more than zero:

`\implies f'(x) > 0`

`\implies 6\left(x+(1)/(2)\right)^2-(39)/(2) > 0`

`\implies 6\left(x+(1)/(2)\right)^2 > (39)/(2)`

`\implies \left(x+(1)/(2)\right)^2 > (39)/(12)`

`\implies \left(x+(1)/(2)\right)^2 > (13)/(4)`

`\textsf{For\;\;$u^n > a$,\;\;if\;$n$\;is\;even\;then\;\;$u < -\sqrt[n]{a}$\;\;or\;\;$u > \sqrt[n]{a}$}.`

Therefore:

`x+(1)/(2) < -\sqrt{(13)/(4)}\implies x < (-1-√(13))/(2)`

`x+(1)/(2) > \sqrt{(13)/(4)}\implies x < (-1+√(13))/(2)`

So the interval on which function f(x) is increasing is:

`\left(-\infty, (-1-√(13))/(2)\right) \cup \left((-1+√(13))/(2), \infty\right)`

Decreasing

To find the interval where f(x) is decreasing, set the differentiated function to less than zero:

`\implies f'(x) < 0`

`\implies 6\left(x+(1)/(2)\right)^2-(39)/(2) < 0`

`\implies \left(x+(1)/(2)\right)^2 < (13)/(4)`

`\textsf{For\;\;$u^n < a$,\;\;if\;$n$\;is\;even\;then\;\;$-\sqrt[n]{a} < u < \sqrt[n]{a}$}.`

Therefore:

`x+(1)/(2) > -\sqrt{(13)/(4)}\implies x > (-1-√(13))/(2)`

`x+(1)/(2) < \sqrt{(13)/(4)}\implies x < (-1+√(13))/(2)`

So the interval on which function f(x) is decreasing is:

`\left((-1-√(13))/(2) < x < (-1+√(13))/(2)\right)`

Part (b)

To find x-coordinates of the local minimum and maximum set the differentiated function to zero and solve for x:

`\implies f'(x)=0`

`\implies 6\left(x+(1)/(2)\right)^2-(39)/(2) =0`

`\implies \left(x+(1)/(2)\right)^2 =(13)/(4)`

`\implies x=(-1-√(13))/(2),\;\;(-1+√(13))/(2)`

To find the y-coordinates of the turning points, substitute the found values of x into the function and solve for y:

`\implies f\left((-1-√(13))/(2)\right)=(19+13√(13))/(2)`

`\implies f\left((-1+√(13))/(2)\right)=(19-13√(13))/(2)`

Therefore:

`\textsf{Minimum}: \quad \left((-1-√(13))/(2),(19+13√(13))/(2)\right) \approx (-2.30, 32.94)`

`\textsf{Maximum}: \quad \left((-1+√(13))/(2),(19-13√(13))/(2)\right) \approx (1.30, -13.94)`

Part (c)

At a point of inflection, f''(x) = 0.

To find the point of inflection, differentiate the function again:

`\implies f''(x)=12x+6`

Set the second derivative to zero and solve for x:

`\implies f''(x)=0`

`\implies 12x+6=0`

`\implies x=-(1)/(2)`

Substitute the found value of x into the original function to the find the y-coordinate of the point of inflection:

`\implies f\left(-(1)/(2)\right)=(19)/(2)`

Therefore, the inflection point is:

`\left(-(1)/(2), (19)/(2)\right)`

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

`\implies f''(x) > 0`

`\implies 12x+6 > 0`

`\implies x > -(1)/(2)`

`\implies \left(-(1)/(2), \infty\right)`

Concave down

`\implies f''(x) < 0`

`\implies 12x+6 > 0`

`\implies x < -(1)/(2)`

`\implies \left(- \infty,-(1)/(2)\right)`