Question

To two decimal places, find the value of k that will make the function f(x) continuous everywhere.f of x equals the quantity 3 times x plus k for x less than or equal to 3 and is equal to k times x squared minus 6 for x greater than 3

Answer 1

Given the function below

`f(x)=(x+5)/(x^2+9x+20)`

Let us check the graph of the function to determine if it is continuos or not.

It can be observed from the graph above that function is continuos at a particular interval

Note that the function f(x) is continous on interval (a,b) if it is continuos at every point.

In other to determine the continuos interval, we would find the domain as shown below:

`\begin{gathered} f(x)=(x+5)/(x^2+9x+20) \\ f(x)=(x+5)/(x^2+4x+5x+20) \\ f(x)=(x+5)/(x(x+4)+5(x+4)) \\ f(x)=(x+5)/((x+4)(x+5)) \end{gathered}`

Let us equate both the numerator and denominator of the function to zero

`\begin{gathered} Numeraor,x+5=0 \\ x=-5 \\ Deno\min ator,(x+4)(x+5)=0 \\ x+4=0,or,x+5=0 \\ x=-4,or,x=-5 \end{gathered}`

Hence, the function is continuous everywhere, when x < -5, or -5 < x < -4, or x > -4