find the intervals on which f(x) is increasing and decreasing along with the local extrema.
f(x)=x^4 + 18x^2
I took the derivative and got:
f'(x)= 4x^3 + 36x
When I set this to zero, I get the imaginary number 3i. I can't get test values for an imaginary numbers, so I can't find the extrema. Does this have no solution?
4x^3+36x=0=4x(x^2+9)
solutions are at 3i, and Zero.
now look at x=0
bobpursley can u check my work?
To find the intervals on which the function f(x) is increasing and decreasing, as well as the local extrema, you correctly took the derivative of f(x) which is f'(x) = 4x^3 + 36x. However, there is a mistake in your calculation.
Setting f'(x) equal to zero, we have:
4x^3 + 36x = 0
Factoring out 4x, we get:
4x(x^2 + 9) = 0
Now, we have two solutions: x = 0 and x^2 + 9 = 0.
For x^2 + 9 = 0, we subtract 9 from both sides and take the square root:
x^2 = -9
x = ± √(-9)
Since the square root of a negative number is imaginary, we conclude that there are no real solutions for x^2 + 9 = 0.
Thus, the only critical point is x = 0.
To determine the intervals of increasing and decreasing, we can look at the sign of the derivative before and after the critical point x = 0.
Consider a value slightly less than 0, for example x = -1:
f'(-1) = 4(-1)^3 + 36(-1) = -4 - 36 = -40
Since the derivative is negative, the function is decreasing for x < 0.
Consider a value slightly greater than 0, for example x = 1:
f'(1) = 4(1)^3 + 36(1) = 4 + 36 = 40
Since the derivative is positive, the function is increasing for x > 0.
Therefore, the function f(x) = x^4 + 18x^2 is decreasing on the interval (-∞, 0) and increasing on the interval (0, ∞).
As for the local extrema, since there are no real solutions for the equation f'(x) = 0, there are no local extrema for the function f(x) = x^4 + 18x^2.