So that equation was definitely correct...
When you expand the equation in the bracket you'll find out that you'll get a^6 + 4a^4 + !6a^2 - 4a^4 - 16a^2 -64. then your final result will be a^6 - 64
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Answer:
hope this helps
Step-by-step explanation:
Answer: the maximum is 25.
Step-by-step explanation: a max/min can occur on the endpoints of a function and critical points of the function's derivative.
f(x)=x^4-x^2+13
f'(x)=4x^3-2x
The critical points of f'(x) occur when f'(x) is zero or undefined. f'(x) is not ever undefined in this case, so we just need to find the x values for when it's zero.
0=4x^3-2x
x=.707, -.707
Now that we have the critical points of f'(x) (.707 and -.707) and endpoints (-1 and 2), we can plug in these x values into the original function to determine its maximum. When you do this you'll find that the greatest y value produced occurs when x=2 and results in a max of 25.
No the points have to be on a cross on the x and y axis
Answer: $775.50
explanation: I just looked it up and another website had the same question but also had the answer
