Question

The function f is given by f(x)=0.1x4−0.5x3−3.3x2+7.7x−1.99. For how many positive values of b does limx→bf(x)=2 ?

Answer 1

Answer:

3

Step-by-step explanation:

First, plot the graph of the function

$f(x)=0.1x^4-0.5x^3-3.3x^2+7.7x-1.99$

and the graph of the line

$y=2$

This line intersects the graph of the function f(x) at 4 different points. So, there are 4 different values for b at which

$\lim \limits_{x \to b} f(x)=2$

Only 3 of them have positive values.

At these positive values, f(x)=2 (because the graph of the function and the graph of the line y=2 intersect at these points), then for these three b with positive values,

$\lim \limits_{x \to b} f(x)=2$

Answer 2

The total number of positive values of b at which $\lim_{x \to b} f(x) = 2$ where,  $f(x)=0.1x^4-0.5x^3-3.3x^2+7.7x-1.99$  is three that is: 0.951 , 1.049 and 8. This can be determine by sketching the graph of y = 2 and function f(x).

Given :

Function -  $f(x)=0.1x^4-0.5x^3-3.3x^2+7.7x-1.99$

Positive values of b when $\lim_{x \to b} f(x) = 2$  can be evaluated by simply sketching the graph of y = 2 and  $f(x)=0.1x^4-0.5x^3-3.3x^2+7.7x-1.99$.

The combined graph of y = 2 and  $f(x)=0.1x^4-0.5x^3-3.3x^2+7.7x-1.99$ is attached below.

The line y = 2 intersects function  $f(x)=0.1x^4-0.5x^3-3.3x^2+7.7x-1.99$ at four different points that is: (-5 , 2) , (0.951 , 2) , (1.049 , 2) and (8 , 2).

Therefore, the total number of positive values of b at which $\lim_{x \to b} f(x) = 2$  is three that is: 0.951 , 1.049 and 8.

For more information, refer the link given below:

https://brainly.com/question/22115261