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lisabon 2012 [21]
3 years ago
12

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
Mathematics
1 answer:
Fiesta28 [93]3 years ago
7 0

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=-\frac{7}{15}

k=-0.47

<u>Hence final answer is -0.47.</u>




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