Given f(x)=x^4(2x^2-15). On what interval(s) is the graph of f concave upwards?
A. (0, sqrt(3))
B. (-sqrt(3), 0)
C. (-sqrt(3), 0) and (0, sqrt(3))
D. (-sqrt(3), sqrt(3))
E. (Negative infinity, -sqrt(3)) and (sqrt(3), infinity)
I got E
looks good to me.
To determine the intervals on which the graph of f is concave upwards, we need to analyze the second derivative of the function.
First, let's find the second derivative of f(x). Since f(x) = x^4(2x^2 - 15), we can apply the product rule:
f'(x) = 4x^3(2x^2 - 15) + x^4(4x) = 8x^5 - 60x^3 + 4x^5 = 12x^5 - 60x^3.
Next, we differentiate f'(x) with respect to x to obtain the second derivative:
f''(x) = d/dx (12x^5 - 60x^3) = 60x^4 - 180x^2.
Now, we need to find the values of x for which f''(x) > 0. These are the intervals on which the graph of f is concave upwards.
To solve f''(x) > 0, let's factor out common terms:
60x^4 - 180x^2 = 60x^2(x^2 - 3).
Setting each factor equal to zero, we can find the critical points:
60x^2 = 0 => x = 0,
x^2 - 3 = 0 => x^2 = 3 => x = ±√3.
Now we can create a number line and test each interval to determine the sign of f''(x). We'll choose a value from each interval and substitute it into f''(x).
Testing the interval (-∞, -√3):
Let's choose x = -4:
f''(-4) = 60(-4)^2 - 180(-4)^2 = 60(16) - 180(16) = -240 + (-2880) = -3120.
Testing the interval (-√3, 0):
Let's choose x = -1:
f''(-1) = 60(-1)^2 - 180(-1)^2 = 60(1) - 180(1) = 60 - 180 = -120.
Testing the interval (0, √3):
Let's choose x = 1:
f''(1) = 60(1)^2 - 180(1)^2 = 60(1) - 180(1) = 60 - 180 = -120.
Testing the interval (√3, ∞):
Let's choose x = 4:
f''(4) = 60(4)^2 - 180(4)^2 = 60(16) - 180(16) = -240 + (-2880) = -3120.
From these tests, we see that f''(x) < 0 for all x in the intervals (-√3, 0) and (0, √3), indicating the graph of f is concave downwards in these intervals.
Therefore, the correct answer is option E: (Negative infinity, -√3) and (√3, infinity).