I am taking a course in calculus and wanted to refresh my memory before the semester starts. And I have been working on drawing graphs from cubic functions. I am not that experienced with LaTex and english is not my first language, so I apologize in advance if some of the question is improperly formulated and some expressions/words are incorrect.
I have this exercise from my old book from high school which goes as follows:
A function G is given by
$G(x)=\frac{1}{3}x^3-x^2+x-\frac{4}{3}$
a) Decide $G'(x)$, and find the monotonic properties for $G$
b) Decide $G''(x)$, and find the curvature for $G$
c) Calculate the turning point and the equation for the tangent of the turning point
d) Draw the graph of $G$ with the tangent of the turning point (this is where I'm having some trouble)
I have solved the task as follows:
a)
$G'(x)=3\frac{1}{3}x^{3-1}-2x^{2-1}+1=x^2-2x+1$
We know from the second quadratic formula that $(a-b)^2=a^2-2ab+b^2$,
thus $G'(x)=(x-1)(x-1)=(x-1)^2$
Monotonic properties:
$(x-1)\leq 0 \,\forall x \in [-\infty,1]$
$(x-1)\geq 0 \,\forall x \in [1,\infty]$
$G'(x)> 0 \,\forall x \in [-1,1]$
We see that $G'(x)$ is growing on the intervall and it only has one nullpoint/saddlepoint, so it should have a shape similar to this:
b)
$G''(x)=(2x-2)=2(x-1)$
$2>0\,\forall x \in [-\infty,\infty]$
$(x-1)\leq 0 \,\forall x \in [-\infty,1]$
$G''(x)\geq 0 \,\forall x \in [1,\infty]$
$G$ curves downwards when $G''(x)<0\,\forall x \in [-\infty,1]$ and upwards when $G''(x)>0\,\forall x \in [1,\infty]$
c)
Formula for tangent:
Let $f$ be double differentiable in a point $x=c$. The tangent $t(x)$ for $f$ in the point $(c, f(c))$ is then given by
$t(x)=f'(c)(x-c)+f(c)$
We know from a) that $G'(x)=0$ when $x=1$, thus $G$ has a turning point in $(1,G(1))=(1,-1)$,
$c=1$, and we get
$t(x)=f'(c)(x-c)+f(c)=f'(1)(x-1)+f(1)=(1-1)^2(x-1)+(-1)=-1$
d)
We know that there's a saddle point in $(1,-1)$.
Then we calculate where the graph crosses the $y-axis$ by setting $x=0$:
$G(0)=\frac{1}{3}(0^3)-(0^2)+0-(4/3)=-\frac{4}{3}\,\rightarrow$ in the point $(0,-\frac{4}{3})$
But when I try to calculate where the graph crosses the x-axis by putting $y=0$, I'm couldn't figure out how to simplify the cubic function.
$\frac{1}{3}x^3-x^2+x-\frac{4}{2}$
$x^2(\frac{1}{3}x-1)+(\frac{4}{3})(\frac{3}{4}x-1)$
from here, the factors in the two terms are not the same, so I'm not sure how to proceed. What am I doing wrong here? How do I find the point where the graph crosses the x-axis?
Best Answer
It's not very obvious how to factor $\frac{1}{3}x^3-x^2+x-\frac{4}{3}$. Multiplying by $3$ we have $x^3-3x^2+3x-4=(x-1)^3-3$. Can you find the root now?