Explanation Given: Consider the function: h ( x ) = 3 x 6 x 2 + 15 x + 6 x = 0 , 3 , − 1 2 _ Formula used: A function f is continuous at x = c if (a) f ( c ) is defined; (b) lim x → c f ( x ) exists; (c) lim x → c f ( x ) = f ( c ) Calculation: First, check the left hand limit and the right hand limit for the given value of x , if values obtained by solving both the limits are equal, then we say that the function is continuous at that point. For x = 0 ; lim x → 0 h ( x ) = lim x → 0 3 x 6 x 2 + 15 x + 6 For left hand limit, we have: lim x → 0 − h ( x ) = lim h → 0 3 ( 0 − h ) 6 ( 0 − h ) 2 + 15 ( 0 − h ) + 6 = 0 For right hand limit, we have: lim x → 0 + h ( x ) = lim h → 0 3 ( 0 + h ) 6 ( 0 + h ) 2 + 15 ( 0 + h ) + 6 = 0 Therefore, lim x → 0 − h ( x ) = lim x → 0 + h ( x ) = g ( 0 ) = 0 Thus, h ( x ) is continuous at x = 0

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ISBN: 9780134773285

Author: MULLINS

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Chapter 11.9, Problem 14E

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To calculate: If the function h(x)=3x6x2+15x+6 is continuous at given values of x.

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If the function h ( x ) = 3 x 6 x 2 + 15 x + 6 is continuous at given values of x .

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