Question - Analyzing the Composition of a Given Function Graph
Solution:
Given the function \( f(x) = 2x^3 - 9x^2 + 12x - 3 \), we are asked to determine its concavity, maximum point, and inflection point.
To find the concavity of the function, we need to compute the second derivative and analyze its sign:
\( f''(x) = \frac{d^2}{dx^2}(2x^3 - 9x^2 + 12x - 3) \)
\( f''(x) = \frac{d}{dx}(6x^2 - 18x + 12) \)
\( f''(x) = 12x - 18 \)
To determine the concavity, we look at the sign of \( f''(x) \).
If \( f''(x) > 0 \), the function is concave up.
If \( f''(x) < 0 \), the function is concave down.
So, the function is concave up when \( 12x - 18 > 0 \) i.e., \( x > \frac{3}{2} \), and concave down when \( x < \frac{3}{2} \).
To find the maximum point, we need to set the first derivative equal to zero and solve for x:
\( f'(x) = \frac{d}{dx}(2x^3 - 9x^2 + 12x - 3) \)
\( f'(x) = 6x^2 - 18x + 12 \)
Setting \( f'(x) = 0 \), we get:
\( 6x^2 - 18x + 12 = 0 \)
Dividing by 6:
\( x^2 - 3x + 2 = 0 \)
Factoring:
\( (x - 2)(x - 1) = 0 \)
Thus, \( x = 1 \) and \( x = 2 \) are critical points.
We must check these points in the second derivative to see if they correspond to maximum points:
\( f''(1) = 12(1) - 18 = -6 < 0 \), so \( x = 1 \) is a local maximum point.
\( f''(2) = 12(2) - 18 = 6 > 0 \), so \( x = 2 \) is not a maximum point, it is a point of inflection since the concavity changes from down to up.
CamTutor
In regards to math, we are professionals.
Get In Touch
Email: camtutor.ai@gmail.com