5 mark for review the first derivative of the function f is defined by f(x)=(x² + 1)sin(3x - 1) for - 1.5 <…

5 mark for review the first derivative of the function f is defined by f(x)=(x² + 1)sin(3x - 1) for - 1.5 < x < 1.5. on which of the following intervals is the graph of f concave up? a (-1.5, - 1.341) and (-0.240, 0.964) b (-1.341, - 0.240) and (0.964, 1.5) c (-0.714, 0.333) and (1.381, 1.5) d (-1.5, - 0.714) and (0.333, 1.381)

5 mark for review the first derivative of the function f is defined by f(x)=(x² + 1)sin(3x - 1) for - 1.5 < x < 1.5. on which of the following intervals is the graph of f concave up? a (-1.5, - 1.341) and (-0.240, 0.964) b (-1.341, - 0.240) and (0.964, 1.5) c (-0.714, 0.333) and (1.381, 1.5) d (-1.5, - 0.714) and (0.333, 1.381)

Answer

Explanation:

Step1: Recall concavity condition

The graph of a function $y = f(x)$ is concave - up when $f''(x)>0$. Given $f'(x)=(x^{2}+1)\sin(3x - 1)$, we use the product rule $(uv)' = u'v+uv'$ to find $f''(x)$. Let $u=x^{2}+1$ and $v = \sin(3x - 1)$. Then $u'=2x$ and $v'=3\cos(3x - 1)$. So $f''(x)=2x\sin(3x - 1)+3(x^{2}+1)\cos(3x - 1)$.

Step2: Use a graphing utility or numerical method

Since solving $f''(x)=0$ analytically is difficult, we can use a graphing calculator or software (e.g., Desmos) to graph $y = f''(x)$ for $-1.5<x<1.5$. We look for the intervals where the graph of $y = f''(x)$ is above the $x -$axis. By graphing $y = f''(x)=2x\sin(3x - 1)+3(x^{2}+1)\cos(3x - 1)$ in the interval $-1.5 < x<1.5$, we find that $f''(x)>0$ on the intervals $(-1.341,-0.240)$ and $(0.964,1.5)$.

Answer:

B. $(-1.341,-0.240)$ and $(0.964,1.5)$