Answer:
A) Increasing: (-0.5, ∞)
Decreasing: (-∞, -2) ∪ (-2, -0.5)
B) x = -¹/₂
C) Concave up: (-∞, -2) ∪ (-1, ∞)
Concave down: (-2, -1)
Points of inflection: (-2, 0) and (-1, -1)
D) See attachments.
Step-by-step explanation:
Given function and derivatives:
[tex]f(x) = x(x+2)^3[/tex]
[tex]f'(x) = (4x+2)(x+2)^2[/tex]
[tex]f''(x) = 12(x+1)(x+2)[/tex]
[tex]\textsf{A function is \textbf{increasing} when the \underline{gradient is positive}}\implies f'(x) > 0[/tex]
[tex]\textsf{A function is \textbf{decreasing} when the \underline{gradient is negative}} \implies f'(x) < 0[/tex]
Increasing function
[tex]\begin{aligned}f'(x) & > 0\\(4x+2)(x+2)^2 & > 0\\2(2x+1)(x+2)^2 & > 0\\(2x+1)(x+2)^2 & > 0\end{aligned}[/tex]
Therefore, the function is increasing on the Interval:
[tex]\left(-\dfrac{1}{2}, \infty \right)[/tex]
Decreasing function
[tex]\begin{aligned}f'(x) & < 0\\(4x+2)(x+2)^2 & < 0\\2(2x+1)(x+2)^2 & < 0\\(2x+1)(x+2)^2 & < 0\end{aligned}[/tex]
Therefore, the function is decreasing on the Interval:
[tex](- \infty, -2) \cup \left(-2,-\dfrac{1}{2} \right)[/tex]
Local minimum/maximum points occur when f'(x) = 0.
[tex]\begin{aligned}f'(x) &= 0\\(4x+2)(x+2)^2 & = 0\\2(2x+1)(x+2)^2 & = 0\\(2x+1)(x+2)^2 &= 0\\\\2x+1&=0 \implies x=-\dfrac{1}{2}\\(x+2)^2&=0 \implies x=-2\end{aligned}[/tex]
[tex]f''\left(-\dfrac{1}{2}\right) = 12\left(-\dfrac{1}{2}+1 \right)\left(-\dfrac{1}{2}+2 \right)=9 > 0\implies \textsf{minimum}[/tex]
[tex]f''\left(-2\right) = 12\left(-2+1 \right)\left(-2+2 \right)=0\implies \textsf{point of inflection}[/tex]
Therefore, the value of x where f(x) has a local minimum is x = -¹/₂.
At a point of inflection, f''(x) = 0.
[tex]\begin{aligned}f''(x) &=0\\12(x+1)(x+2)&=0\\(x+1)(x+2)&=0\\\\x+1&=0 \implies x=-1\\x+2&=0 \implies x=-2\end{aligned}[/tex]
Substitute the found values of x into the original function to the find the y-coordinates of the points of inflection:
[tex]\implies f(-1) = -1(-1+2)^3=-1[/tex]
[tex]\implies f(-2) = -2(-2+2)^3=0[/tex]
Therefore, the inflection points are:
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
[tex]\implies f''(x) > 0[/tex]
[tex]\implies 12(x+1)(x+2) > 0[/tex]
[tex]\implies (x+1)(x+2) > 0[/tex]
Therefore, the function is concave up at:
[tex]\left(-\infty,-2\right) \cup \left(-1,\infty \right)[/tex]
Concave down
[tex]\implies f''(x) < 0[/tex]
[tex]\implies 12(x+1)(x+2) < 0[/tex]
[tex]\implies (x+1)(x+2) < 0[/tex]
Therefore, the function is concave down at:
[tex](-2,-1)[/tex]
See attachment 1 for how to sketch the graph using the information from parts A-C.
See attachment 2 for the final sketch of the graph.
