Question

Sketch one graph of a single function f that satisfies all of the conditions below

a. domain (0, infinity)
b. limit as x -> 0+ f(x) = infinity
c. limit as x-> infinity f(x) = 2
d. f'(x) < 0 on the interval (0,3)
e. f'(x) > 0 on the interval (3, infinity)
f. f'(3) = 0
g. f''(x) > 0 on the interval (0,6)
h. f''(x) < 0 on the interval (6, infinity)

Answer 1

Answer on Question #48103 – Math – Calculus

Sketch one graph of a single function $f$ that satisfies all of the conditions below:

a. domain is $(0, \infty)$;

b. $\lim_{x\to 0_+}f(x) = \infty$;

c. $\lim_{x\to \infty}f(x) = 2$;

d. $f^{\prime}(x) < 0$ on the interval $(0,3)$;

e. $f^{\prime}(x) > 0$ on the interval $(3,\infty)$;

f. $f^{\prime}(3) = 0$;

g. $f^{\prime \prime}(x) > 0$ on the interval $(0,6)$

h. $f^{\prime \prime}(x) < 0$ on the interval $(6,\infty)$.

Solution.

By a, domain of function $f(x)$ is $(0, \infty)$, therefore $x > 0$.

By d, $\forall x\in (0,3)\colon f^{\prime}(x) < 0\Rightarrow f$ decreases on $(0,3)$. By b, $\lim_{x\to 0_+}f(x) = +\infty$.

By c and e,

$$\left\{ \begin{array}{c} \lim _ {x \to \infty} f (x) = 2 \\ \forall x \in (3, + \infty): f ^ {\prime} (x) > 0 \end{array} \right. \Rightarrow \forall x \in (3, \infty): f (x) < 2; f \text{ increases on } (3, \infty).$$

Take into account d, e, f, so $f'(x) < 0$ on $(0,3)$; $f'(x) > 0$ on $(3,\infty)$; $f'(3) = 0 \Rightarrow x = 3$ is a point of local minimum (in this problem it will be global minimum). Note that by g and h, $f''(x) > 0$ on $(0,6)$, $f''(x) < 0$ on $(6,\infty) \Rightarrow$ the graph of function is convex downward on $(0,6)$, the graph of function is convex upward on $(6,\infty)$.

So, the graph of $f(x)$ can be like this:

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