Let-f-x-3x-2-1-x-lt-0-cx-d-0-x-1-x-8-x-gt-1-find-the-value-of-c-amp-d-such-that-f-x-continous-everywhere-

Question Number 140073 by EDWIN88 last updated on 04/May/21

Let f (x) = {\begin{cases} {3 x}^{2} - 1; x < 0 \\ cx + d; 0 ⩽ x ⩽ 1 \\ \sqrt{x + 8}; x > 1 \end{cases}

find the value of c & d such that f (x) continous

everywhere

Answered by bobhans last updated on 04/May/21

since \lim_{x \to 0^{-}} {3 x}^{2} - 1 = - 1 the value of

cx + d at x = 0 must be - 1 that is d = - 1

since \lim_{x \to 1^{+}} \sqrt{x + 8} = 3 the value of cx + d at x = 1

must be 3 that is 3 = c (1) - 1, c = 4

∴ {\begin{cases} c = 4 \\ d = - 1 \end{cases}

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