The function f(x) is defined by f(x)=\left\{\begin{array}{c}x^{2}+a x+b, 0 \leq x<2 \\ 3 x+2,2 \leq x \leq 4 \\ 2 a x+5 b, 4<x \leq 8\end{array}\right.If f is continuous on [0, 8], find the values of a and b.
The function f(x) is defined by f(x)=\left\{\begin{array}{c}x^{2}+a x+b, 0 \leq x<2 \\ 3 x+2,2 \leq x \leq 4 \\ 2 a x+5 b, 4<x \leq 8\end{array}\right.If f is continuous on [0, 8], find the values of a and b.

A real function f is said to be continuous at x = c, where c is any point in the domain of f if

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h is a very small positive number. i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

A function is continuous at x = c if

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From equation 1,  f(x) is changing its expression at x = 2

Given, f (x) is continuous everywhere

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4+2a + b = 8

∴ 2a + b = 4

∴ b = 4 – 2a …………… equation 2

Also from equation 1, it is clear that f(x) is also changing its expression at x = 4

Given, f (x) is continuous everywhere

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∴ 8a + 5b = 14 ……………….Equation 3

Putting value of a from equation 2 to equation 3

    \[\begin{array}{*{35}{l}} \therefore ~8a\text{ }+\text{ }5\left( 4-2a \right)\text{ }=\text{ }14  \\ \Rightarrow ~2a\text{ }=\text{ }6  \\ \therefore ~a\text{ }=\text{ }6/2  \\ =\text{ }3  \\ \therefore ~b\text{ }=\text{ }4\text{ }-\text{ }2\times 3\text{ }=-\text{ }2  \\ Thus,\text{ }a\text{ }=\text{ }3\text{ }and\text{ }b\text{ }=-\text{ }2  \\ \end{array}\]