Find the points at which the function $f$ given by $f(x)=(x-4)^4(x+1)^3$ has local maxima, minima, point of inflection.

Question

Find the points at which the function $f$ given by $f(x)=(x-4)^4(x+1)^3$ has local maxima, minima, point of inflection.

By differentiating I found the critical points to be at $-1,\frac87,4$. However, I could not do anything more. Please help.

Answer 1

$$f'(x)=4(x-4)^3(x+1)^3+23(x+1)^2(x-4)^4=(x+1)^2(x-4)^3(7x-8),$$ which says that $x_{min}=4$ because sing of $f'$ changes from $-$ to $+$, $x_{max}=\frac{8}{7}$ because sing of $f'$ changes from $+$ to $-$and $x=-1$ gives an inflection point because sing of $f'$ does not change.

But you need also to calculate $f''(x).$

Answer 2

Note that $f(x)$ is a polynomial with $\deg(f)=7$ and principal coefficient $1$. Then $$\lim_{x\to\pm\infty}f(x)=\pm\infty$$

Now, because $mult(4,f)\equiv 0 \mod 2$ and $mult(-1,f)\equiv 1 \mod 2$ we have that

• $f(x)>0$ when $x\in(-1,4)\cup(4,+\infty)$
• $f(x)<0$ when $x\in(-\infty,-1)$

Whit this, we can conclude that:

• $x=4$ is a local minima (because $f(x)>0$ around that)
• $x=-1$ is inflection (check the second derivate)
• $x=\frac{8}{7}$ is a local maxima (because $f(x)>0$ in $(-1,4)$)

Answer 3

Note that $$f’(x) = (x-4)^4(3(x+1)^2) + (x+1)^3(4(x-4)^3) = (x-4)^3(x+1)^2[3x-12+4x+4] = (x-4)^3(x+1)^2[7x-8]$$ and $$f’’(x)= 7\left(x-4\right)^3\left(x+1\right)^2+3\left(x- 4\right)^2\left(x+1\right)^2\left(7x-8\right)+2\left(x-4\right)^3\left(x+1\right)\left(7x-8\right) = 6\left(x-4\right)^2\left(x+1\right)\left(7x^2-16x+2\right)$$

Now, note that if $f’’(x)\lvert_{x_1} <0$, where $f’(x_1)=0$, then $x_1$ is a maxima, the opposite holds true for a point which is a minima. To check inflection points, follow the procedure here.