which of the following is true about the function g(x)= 4x^2 - 3x^4?
a)g is decreasing for x<1
b) g has a relative extreme value at (0,0)
c)the graph of g is concave up for all x<0
g' = 8x-12x^3 = 4x(2-x^2)
g" = 8-36x^2
g' changes sign in x<1, so not (a)
g'=0 and g"≠0 at x=0, so (b)
g" changes sign, so not (c)
so, (b) is the only true statement
the answer is a
g'(1/2) = 4(1/2)(2-1/4) = 2(7/4) = 7/2
So, g is increasing at x = 1/2
Have you looked at the graph?
http://www.wolframalpha.com/input/?i=4x^2+-+3x^4
Clearly only (b) is true
To determine which of the given statements is true about the function g(x) = 4x^2 - 3x^4, we will analyze each statement individually.
a) g is decreasing for x < 1:
To determine the increasing or decreasing nature of a function, we need to find the first derivative of the function. Let's find the derivative of g(x):
g'(x) = d/dx(4x^2 - 3x^4)
= 8x - 12x^3
To find the intervals where g is increasing or decreasing, we can set g'(x) = 0 and solve for x:
8x - 12x^3 = 0
4x(2 - 3x^2) = 0
This equation has two critical points:
1) x = 0
2) x = sqrt(2/3)
Now, let's analyze the intervals using a number line:
Interval (-∞, 0):
Plug in a test value, such as x = -1, into g'(x) = 8x - 12x^3:
g'(-1) = 8(-1) - 12(-1)^3
= -8 - 12(-1)
= -8 + 12
= 4
Since g'(-1) > 0, g is increasing in this interval, not decreasing.
Interval (0, √(2/3)):
Plug in a test value, such as x = 0.5, into g'(x) = 8x - 12x^3:
g'(0.5) = 8(0.5) - 12(0.5)^3
= 4 - 12(0.125)
= 4 - 1.5
= 2.5
Since g'(0.5) > 0, g is increasing in this interval, not decreasing.
Interval (√(2/3), ∞):
Plug in a test value, such as x = 2, into g'(x) = 8x - 12x^3:
g'(2) = 8(2) - 12(2)^3
= 16 - 12(8)
= 16 - 96
= -80
Since g'(2) < 0, g is decreasing in this interval.
Based on the analysis, g is only decreasing for x > sqrt(2/3), so statement (a) is false.
b) g has a relative extreme value at (0,0):
To find the relative extreme values, we need to find the second derivative of g(x). Let's find the derivative:
g''(x) = d^2/dx^2(4x^2 - 3x^4)
= 8 - 36x^2
Now let's evaluate g''(0) to determine if there is a relative extreme value at (0, 0):
g''(0) = 8 - 36(0)^2
= 8
Since g''(0) > 0, the function g(x) = 4x^2 - 3x^4 has a relative minimum at (0, 0). Hence, statement (b) is true.
c) The graph of g is concave up for all x < 0:
To determine the concavity of the function, we need to analyze the sign of the second derivative, g''(x).
For any value of x less than zero, g''(x) = 8 - 36x^2 will always be positive since x^2 is non-negative. Therefore, the graph of g is concave up for all x < 0. Hence, statement (c) is true.
In conclusion:
The correct statements are:
- g has a relative extreme value at (0, 0)
- The graph of g is concave up for all x < 0.