Determine a function based on following criteria

Question

I need to create a function with these rules:

$1)$ $f(3)=8$

$2)$ The function is continuous at $x=3$

$3)$ Concave down on $(-\infty,3)$ and concave up on $(3, \infty)$

$4)$ Slope of tangent line cannot equal zero at $x=3$

How can I manipulate a function so that the slope of the tangent line IS NOT zero but also follows the rules above?

I've tried functions like these for example:

$y=(3x-9)^5$

$y=(x-3)^3$

As you can see if $x=3$ then it won't work.

Answer 1

Notice that the second derivative has a zero at $x=3:$

$$f''(3)=0\implies f''(x)=x-3$$

$$\implies f'(x)=\int f''(x)dx=\frac12x^2-3x+c$$

$$f'(3)\ne0\implies c\ne\frac92$$

Just as an example, we will choose $c=0$.

$$f(x)=\int f'(x)dx=\frac16x^3-\frac32x^2+d$$

$$f(3)=8\implies d=17$$

Thus, one such solution is given by

$$f(x)=\frac16x^3-\frac32x^2+17$$

Answer 2

If f is piecewise, you can do the following:

$ \;f(x) = \begin{cases} 2^x & \text{if } x\ge3\\ -(x-3)^2+8 &\text{if } x\lt0 \end{cases} $

although this may not have continuous first derivative (so tangent line may not even be defined), but you only mentioned that you need it not to be $0$ so this meets the criteria.

Answer 3

Let $y=8-\arctan(x-3){}{}{}{}$.