# How to draw graph of this function

Given f(x) is a function where its 2nd derivatives f''(x) as shown in the graph below.

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Draw graph for f of possible statement below.

a) f increasing at [-1 , 1]

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-1                 1


b) f have minimum point at x = 0

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-1                 1


c) f decreasing at [-1 , 1]

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-1                 1

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Hint: Your function is of the form $f''(x)=y_0-mx$. Integrate this twice, and you will get a cubic with two adjustable parameters. You should be able to choose those to make the behaviors you want.

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You have to find the antiderivative of the derivatives. The first integration will give you f' and the second one f itself. You can use antiderivative formulas for finding them.

$f''(x)=a.x+b$ then

$f'(x)= a.x^2/2 + b.x + K$

and integrating it one more,

$f(x)=a/6.x^3+b/2.x^2+K.x+K'$

where K and K' are arbitrary constant since deviating doesn't take count of the constant. You can then use a simple function graphing app like Geogebra or even WolframAlpha to plot the function.

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We are given that $f''(x)=mx+b$ for some constants $m$ and $b$. It can be seen that $f''(0)=b=y_0$ is the y-intercept and $m=\frac{y_2-y_1}{x_2-x_1}$ is the slope. Hence both constants are determined by the intercepts (or any other pair of points).

Now, we integrate it twice:

$$f'(x)=\frac{m}{2}x^2+bx+A,\quad A\in\mathbb{R}$$ $$f(x)=\frac{m}{6}x^3+\frac{b}{2}x^2+Ax+B,\quad B\in\mathbb{R}$$

Finally, we consider the different cases:

\begin{align*}&\textbf{a})\quad f(x) \text{ increasing at } x\in[-1,1] \iff f'(x)>0 \text{ at } x\in[-1,1] \implies f'(0)=A>0\\ &\textbf{b})\quad f(x) \text{ attains a minimum at } x=0 \implies f'(0)=0 \,\text{ & }\, f''(0)\ge0 \implies f'(0)=A=0 \,\text{ & }\, f''(0)=b\ge0\\ &\textbf{c})\quad f(x) \text{ decreasing at } x\in[-1,1] \iff f'(x)<0 \text{ at } x\in[-1,1] \implies f'(0)=A<0\end{align*}

It remains to choose $A$ and $B$ such that the actual condition is satisfied. Observe that some implications only work in one direction, so manual checks in the opposite direction may be needed. Also note that there are no restrictions on $B$.

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It would be more traditional to use $m$ for the slope and $b$ or $y_0$ for the intercept. This is correct, but people might read it incorrectly. – Ross Millikan Apr 19 '13 at 20:30
@Ross Millikan Thank you for the input. – Librecoin Apr 19 '13 at 20:47
actually the question contain few constant and more variables. can we really draw the graph for 3 statement ask? is it we need to assume to plug-in any constant in the function attain by doing the integral. if we really can draw the graph, can anyone describe to me how the graph will look like for the question a, b and c. is the graph will be a curve, a straight line, a concave up curve or concave down curve or else. – dgnamu Apr 20 '13 at 2:38
Say $m=2$ and $b=1$. Then $f(x)=\frac{1}{3}x^3+\frac{1}{2}x^2+Ax+B$ and $f'(x)=x^2+x+A$. Let $I=[-1,1]$. As $\min_{x\in I} \{f'(x)-A\}=-\frac{1}{4}$ and $\max_{x\in I} \{f'(x)-A\}=2$, it follows that $A>\frac{1}{4}$ in a) and $A<-2$ in c). $A=0$ satisfies b). The constant term $B$, which does not contribute to how the function changes, does not matter. – Librecoin Apr 20 '13 at 14:06