Answer:
Step-by-step explanation:
This equation looks complicated.We have to make it easier
let's say x^2/3 = t and x^4/3 = t^2
t^2-10t+21=0 [ we can factorize this equation as a (t-3)(t-7) ]
(t-3)(t-7)=0 [ that means , t can be 3 or 7 ]
But don't forget we have to find x not t so,
t=x^2/3=3 ∛ x^2 = 3 x^2 = 9 x=3 or x= -3
t=x^2/3=7 ∛x^2 = 7 x^2 = 343 x ~18.5 or x ~ -18.5
Answer:
d. 9.95h + 5 < 125
Step-by-step explanation:
i hope this helps :)
Hello there!
To find the increasing intervals for this graph just based on the equation, we should find the turning points first.
Take the derivative of f(x)...
f(x)=-x²+3x+8
f'(x)=-2x+3
Set f'(x) equal to 0...
0=-2x+3
-3=-2x
3/2=x
This means that the x-value of our turning point is 3/2. Now we need to analyze the equation to figure out the end behavior of this graph as x approaches infinity and negative infinity.
Since the leading coefficient is -1, as x approaches ∞, f(x) approaches -∞ Because the exponent of the leading term is even, the end behavior of f(x) as x approaches -∞ is also -∞.
This means that the interval by which this parabola is increasing is...
(-∞,3/2)
PLEASE DON'T include 3/2 on the increasing interval because it's a turning point. The slope of the tangent line to the turning point is 0 so the graph isn't increasing OR decreasing at this point.
I really hope this helps!
Best wishes :)
