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MrRa [10]
1 year ago
7

Find the Value of k that will make the function f(x) continuous everywhere.​

Mathematics
1 answer:
cricket20 [7]1 year ago
6 0

well, if that function f(x) were to be continuos on all subfunctions, that means that whatever value 7x + k has when x = 2, meets or matches the value that kx² - 6 has when x = 2 as well, so then 7x + k = kx² - 6 when f(2)

f(x)= \begin{cases} 7x+k,&x\leqslant 2\\ kx^2-6&x > 2 \end{cases}\qquad \qquad f(2)= \begin{cases} 7(2)+k,&x\leqslant 2\\ k(2)^2-6&x > 2 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ 7(2)+k~~ = ~~k(2)^2-6\implies 14+k~~ = ~~4k-6 \\\\\\ 14~~ = ~~3k-6\implies 20~~ = ~~3k\implies \cfrac{20}{3}=k

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sashaice [31]

Hey there!

Assuming you meant

5r^2 - 12 = 68

If so, follow these steps so it can be a little bit easier to solve

Firstly, we have to add by 12 on each of your sides

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Now, we have divide by 5 on each of your sides

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