Let g(x)= square root49x2 − 9The function g is increasing on the following interval(s):  (−∞, ∞)(−∞, a)    (−∞, a](a, ∞)[a, ∞)(−∞, a) ∪ (b, ∞)(−∞, a] ∪ [b, ∞)(a, b)(a, b][a, b)[a, b]None a = b = g is decreasing on the following interval(s):(−∞, ∞)(−∞, c)    (−∞, c](c, ∞)[c, ∞)(−∞, c) ∪ (d, ∞)(−∞, c] ∪ [d, ∞)(c, d)(c, d][c, d)[c, d]Nonec = d =

Question
Asked Oct 15, 2019
39 views
Let g(x)=
 

square root

49x2 − 9


The function g is increasing on the following interval(s):

 

 
(−∞, ∞)
(−∞, a)
    
(−∞, a]
(a, ∞)
[a, ∞)
(−∞, a) ∪ (b, ∞)
(−∞, a] ∪ [b, ∞)
(a, b)
(a, b]
[a, b)
[a, b]
None

 

a = 
b = 


g is decreasing on the following interval(s):

(−∞, ∞)
(−∞, c)
    
(−∞, c]
(c, ∞)
[c, ∞)
(−∞, c) ∪ (d, ∞)
(−∞, c] ∪ [d, ∞)
(c, d)
(c, d]
[c, d)
[c, d]
None



c
d

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Expert Answer

Step 1

Find the first derivative of the given function as follows.

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Image Transcriptionclose

g(x)/49x2-9 d 1 (49к? -9) g(x) 249x29 dr 1 -(98x) 2/49x2 -9 49x 49x2 -9

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Step 2

For g(x) to be increasing g’(x) has to be...

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49x g(x) 49x2-9 49x >0 0 49 x 0 Since the domain of V49x2 - 9 is {x: 49x2-9 >0} 49x2 90 49x29 9 49 3 x or x < 7 3 _ The function g(x) is increasing on the interval (a,c0) with a 7

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Math

Calculus

Functions