Question

17-22: Find the intervals on which \(f\) is concave upward or concave downward, and find the inflection points of \(f\).

Short answer

The function \(f\) is concave upward on \(\left( {\frac{3}{2},\infty } \right)\) and concave downward on \(\left( { - \infty ,\frac{3}{2}} \right)\). The inflection point is \(\left( {\frac{3}{2},\frac{3}{2}} \right)\).

Step-by-step solution

Write the concavity test

If the function\(f''\left( x \right) > 0\)on an interval, then the graph of the function\(f\)is said to beconcave upwardon that interval.

If the function \(f''\left( x \right) < 0\) on an interval, then the graph of the function \(f\) is said to be concave downward on that interval.

Step 2: Find the solutions of the function

Consider the function \(f\left( x \right) = 2{x^3} - 9{x^2} + 12x - 3\).

Differentiate the function w.r.t. \(x\) twice.

\(\begin{aligned}{c}f'\left( x \right) &= \frac{d}{{dx}}\left( {2{x^3} - 9{x^2} + 12x - 3} \right)\\ &= \frac{d}{{dx}}\left( {2{x^3}} \right) - 9\frac{d}{{dx}}\left( {{x^2}} \right) + 12\frac{d}{{dx}}\left( x \right) + \frac{d}{{dx}}\left( { - 3} \right)\\ &= 6{x^2} - 18x + 12\\f''\left( x \right) &= \frac{d}{{dx}}\left( {6{x^2} - 18x + 12} \right)\\ &= \frac{d}{{dx}}\left( {6{x^2}} \right) - 18\frac{d}{{dx}}\left( x \right) + \frac{d}{{dx}}\left( {12} \right)\\ &= 12x - 18\\ &= 12\left( {x - \frac{3}{2}} \right)\end{aligned}\)

Since \(f''\left( x \right) = 0\) then find the solutions of the functions.

\(\begin{aligned}{c}12\left( {x - \frac{3}{2}} \right) &= 0\\x &= \frac{3}{2}\end{aligned}\)

If \(f''\left( x \right) > 0\) then \(x > \frac{3}{2}\) and if \(f''\left( x \right) < 0\) then \(x < \frac{3}{2}\).

Thus, \(f\) isconcave upward on \(\left( {\frac{3}{2},\infty } \right)\) and concave downward on \(\left( { - \infty ,\frac{3}{2}} \right)\).

Step 3: Find the inflection point

Consider the function\(f\left( x \right) = 2{x^3} - 9{x^2} + 12x - 3\).

As the inflection point is at \(\left( {\frac{3}{2},f\left( {\frac{3}{2}} \right)} \right)\). Find \(f\left( 1 \right)\).

\(\begin{aligned}{c}f\left( {\frac{3}{2}} \right) &= 2{\left( {\frac{3}{2}} \right)^3} - 9{\left( {\frac{3}{2}} \right)^2} + 12\left( {\frac{3}{2}} \right) - 3\\ &= 2\left( {\frac{{27}}{8}} \right) - 9\left( {\frac{9}{4}} \right) + 18 - 3\\ &= \frac{3}{2}\end{aligned}\)

Thus, the inflection point is \(\left( {\frac{3}{2},\frac{3}{2}} \right)\).