Respuesta :
The function y = x⁴ + 7 is decreasing on (-∞ , 0) and increasing on (0 , ∞) .
In the question ,
it is given that ,
the function is y = x⁴ + 7 ,
we have to show that function y is decreasing on (-∞ , 0) and increasing on (0 , ∞) .
differentiating the function y = x⁴ + 7 , with respect to x ,
we get ,
y'(x) = d/dx(x⁴ + 7) = 4x³
putting y'(x) = 0 , then 4x³ = 0 ;
x³ = 0 that means , x = 0 .
thus , the critical number is ⇒ x = 0 .
y'(-1) = 4(-1)³ = -4 So , it is decreasing
y'(1) = 4(1)³ = 4 So , it is increasing .
Therefore , the function y is decreasing in (-∞ , 0) and increasing in (0 , ∞) .
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