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find the constant at such that the function is continous everywhere f(x)={(3x^(2),x>=1),(ax+5,x<1):}

User Psamwel
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To find the constant 'a' for the given piecewise function to ensure it's continuous, we must ensure the two parts of the function are equal at x = 1.

This function is divided into two parts:

- f(x) = 3x^2 for values x >= 1
- f(x) = ax + 5 for values x < 1

The point of intersection for these two functions is at the limiting boundary, which is x = 1. Hence, for the function to be continuous, the output of both these parts must be equal at x = 1.

Let's calculate the output for each function at x = 1:

- For f(x) = 3x^2, the output at x = 1 is then f(1) = 3*1^2 = 3
- For f(x) = ax + 5, the output at x = 1 is simply a*1 + 5 = a + 5

For the function to be continuous, these two outputs must be equal. So, we form the equation 3 = a + 5.

If we solve this equation for 'a' by subtracting 5 from both sides, we obtain a = 3 - 5, which simplifies to a = -2.

So, the constant 'a' that makes this function continuous everywhere is -2.

User Ran Davidovitz
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