Question

consider the function $f(x) = 12x^5 + 45x^4 - 200x^3 + 6.$$f(x)$ has inflection points at (reading from left to right) $x$ = $d$, $e$, and $f$where $d$ isand $e$ isand $f$ isfor each of the following intervals, tell whether $f(x)$ is concave up or concave down.$(-infty, d)$: select an answer$(d, e)$: select an answer$(e, f)$: select an answer$(f, infty)$: select an answerquestion help: video message instructor

Explanation:

Step1: Find the second derivative of \( f(x) \)

First, find the first derivative \( f'(x) \):
\( f(x) = 12x^5 + 45x^4 - 200x^3 + 6 \)
\( f'(x) = 60x^4 + 180x^3 - 600x^2 \)

Then, find the second derivative \( f''(x) \) by differentiating \( f'(x) \):
\( f''(x) = 240x^3 + 540x^2 - 1200x \)
Factor out the greatest common factor, which is \( 60x \):
\( f''(x) = 60x(4x^2 + 9x - 20) \)

Now, factor the quadratic \( 4x^2 + 9x - 20 \). We need two numbers that multiply to \( 4\times(-20)= -80 \) and add to \( 9 \). The numbers are \( 16 \) and \( -5 \).
So, \( 4x^2 + 9x - 20 = 4x^2 + 16x - 5x - 20 = 4x(x + 4) - 5(x + 4) = (4x - 5)(x + 4) \)

Thus, the second derivative is \( f''(x) = 60x(4x - 5)(x + 4) \)

Step2: Find the critical points of \( f''(x) \) (inflection points candidates)

Set \( f''(x) = 0 \):
\( 60x(4x - 5)(x + 4) = 0 \)
Using the zero - product property, we have:
\( 60x = 0\) or \(4x - 5 = 0\) or \(x + 4 = 0\)

• For \( 60x = 0\), we get \( x = 0 \)
• For \( 4x - 5 = 0\), we get \( x=\frac{5}{4}=1.25 \)
• For \( x + 4 = 0\), we get \( x=-4 \)

Now, we need to check if the concavity changes at these points. We can use test intervals.

First, order the critical points from left to right: \( x=-4 \) (D), \( x = 0 \) (E), \( x=\frac{5}{4} \) (F)

Step3: Determine concavity on each interval

Interval \( (-\infty, - 4) \)

Pick a test point, say \( x=-5 \)
\( f''(-5)=60\times(-5)\times(4\times(-5)-5)\times(-5 + 4)\)
\(=60\times(-5)\times(-20 - 5)\times(-1)\)
\(=60\times(-5)\times(-25)\times(-1)\)
\(=60\times(-125)\)
\(=-7500<0\)
So, \( f(x) \) is concave down on \( (-\infty, - 4) \)

Interval \( (-4,0) \)

Pick a test point, say \( x = - 1 \)
\( f''(-1)=60\times(-1)\times(4\times(-1)-5)\times(-1 + 4)\)
\(=60\times(-1)\times(-4 - 5)\times(3)\)
\(=60\times(-1)\times(-9)\times(3)\)
\(=60\times27\)
\(=1620>0\)
So, \( f(x) \) is concave up on \( (-4,0) \)

Interval \( (0,\frac{5}{4}) \)

Pick a test point, say \( x = 0.5 \)
\( f''(0.5)=60\times(0.5)\times(4\times0.5 - 5)\times(0.5 + 4)\)
\(=30\times(2 - 5)\times(4.5)\)
\(=30\times(-3)\times4.5\)
\(=30\times(-13.5)\)
\(=-405<0\)
So, \( f(x) \) is concave down on \( (0,\frac{5}{4}) \)

Interval \( (\frac{5}{4},\infty) \)

Pick a test point, say \( x = 2 \)
\( f''(2)=60\times(2)\times(4\times2 - 5)\times(2 + 4)\)
\(=120\times(8 - 5)\times6\)
\(=120\times3\times6\)
\(=2160>0\)
So, \( f(x) \) is concave up on \( (\frac{5}{4},\infty) \)

Answer:

• \( D=-4 \), \( E = 0 \), \( F=\frac{5}{4}=1.25 \)
• \( (-\infty,D) \): Concave Down
• \( (D,E) \): Concave Up
• \( (E,F) \): Concave Down
• \( (F,\infty) \): Concave Up