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 3x + k for x less than or equal to -4 and is equal to kx^2 - 5 for x greater than -4

11.00
-2.47
-0.47
None of these

Answer 1

Given function is

`f(x)=\left\{\begin{matrix}3x+k & x\leq -4 \\ kx^2-5 & x> -4\end{matrix}\right.`

now we need to find the value of k such that function f(x) continuous everywhere.

We know that any function f(x) is continuous at point x=a if left hand limit and right hand limits at the point x=a are equal.

So we just need to find both left and right hand limits then set equal to each other to find the value of k

To find the left hand limit (LHD) we plug x=-4 into 3x+k

so LHD= 3(-4)+k

To find the Right hand limit (RHD) we plug x=-4 into

`kx^2-5`

so RHD=
`k(-4)^2-5`

Now set both equal

`k(-4)^2-5=3(-4)+k`

`16k-5=-12+k`

`16k-k=-12+5`

`15k=-7`

`k=-(7)/(15)`

k=-0.47

Hence final answer is -0.47.