Suppose f(x) = x ^ 4 – 4x ^ 2 + 6, and g(x) = 3x ^ 3 – 8x. Prove, via the Intermediate
Value Theorem, that the functions intersect at least twice between x = –2 and x = 4.
To prove that the functions f(x) and g(x) intersect at least twice between x = -2 and x = 4 using the Intermediate Value Theorem, we need to show that there are two values, say a and b, such that (a) and f(b) = g(b).
First, let's find the values of f(x) and g(x) at -2 and 4:
f(-2) = (-2)^4 - 4(-2)^2 + 6 = 16 - 16 + 6 = 6
g(-2) = 3(-2)^3 - 8(-2) = -24 + 16 = -8
f(4) = (4)^4 - 4(4)^2 + 6 = 256 - 64 + 6 = 198
g(4) = 3(4)^3 - 8(4) = 192 - 32 = 160
Now, we need to find two values, say c and d, such that f(c) = g(c) and f(d) = g(d) between x = -2 and x = 4. We can use the IVT by finding intervals where f(x) and g(x) have opposite signs.
Between the interval -2 and 0:
f(-2) = 6 (positive)
g(-2) = -8 (negative)
Between the interval 0 and 2:
f(0) = (0)^4 - 4(0)^2 + 6 = 6 (positive)
g(0) = 0 (positive)
Between the interval 2 and 4:
f(2) = (2)^4 - 4(2)^2 + 6 = 6 (positive)
g(2) = 3(2)^3 - 8(2) = 16 (positive)
From the above analysis, we can conclude that f(x) and g(x) have different signs in the intervals -2 and 0, and in the intervals 2 and 4. This means that the functions must intersect at least once in each of these intervals, thus intersecting at least twice between x = -2 and x = 4.
To prove that the functions f(x) = x^4 - 4x^2 + 6 and g(x) = 3x^3 - 8x intersect at least twice between x = -2 and x = 4, we can use the Intermediate Value Theorem.
The Intermediate Value Theorem states that if a function is continuous on a closed interval [a, b], and takes on values f(a) and f(b) with opposite signs, then there exists at least one value c in the interval (a, b) where f(c) equals zero (i.e., the function crosses the x-axis).
Let's divide the interval [-2, 4] into three subintervals: [-2, 0], [0, 2], and [2, 4]. We will determine if the functions f(x) and g(x) change sign within each subinterval.
1. Subinterval [-2, 0]:
For f(x), substitute x = -2 and x = 0:
f(-2) = (-2)^4 - 4(-2)^2 + 6 = 16 - 16 + 6 = 6
f(0) = (0)^4 - 4(0)^2 + 6 = 0 + 0 + 6 = 6
Since f(-2) = 6 and f(0) = 6, the function f(x) does not change sign within the subinterval [-2, 0].
For g(x), substitute x = -2 and x = 0:
g(-2) = 3(-2)^3 - 8(-2) = -24 + 16 = -8
g(0) = 3(0)^3 - 8(0) = 0
Since g(-2) = -8 and g(0) = 0, the function g(x) changes sign from negative to nonnegative within the subinterval [-2, 0].
2. Subinterval [0, 2]:
For f(x), substitute x = 0 and x = 2:
f(0) = (0)^4 - 4(0)^2 + 6 = 0 + 0 + 6 = 6
f(2) = (2)^4 - 4(2)^2 + 6 = 16 - 16 + 6 = 6
Since f(0) = 6 and f(2) = 6, the function f(x) does not change sign within the subinterval [0, 2].
For g(x), substitute x = 0 and x = 2:
g(0) = 3(0)^3 - 8(0) = 0
g(2) = 3(2)^3 - 8(2) = 24 - 16 = 8
Since g(0) = 0 and g(2) = 8, the function g(x) changes sign from nonnegative to positive within the subinterval [0, 2].
3. Subinterval [2, 4]:
For f(x), substitute x = 2 and x = 4:
f(2) = (2)^4 - 4(2)^2 + 6 = 16 - 16 + 6 = 6
f(4) = (4)^4 - 4(4)^2 + 6 = 256 - 64 + 6 = 198
Since f(2) = 6 and f(4) = 198, the function f(x) changes sign from nonnegative to positive within the subinterval [2, 4].
For g(x), substitute x = 2 and x = 4:
g(2) = 3(2)^3 - 8(2) = 24 - 16 = 8
g(4) = 3(4)^3 - 8(4) = 192 - 32 = 160
Since g(2) = 8 and g(4) = 160, the function g(x) changes sign from positive to positive within the subinterval [2, 4].
Based on the analysis above, we can see that the functions f(x) and g(x) change sign within the subintervals [-2, 0], [0, 2], and [2, 4]. This means that both functions cross the x-axis within each of these intervals. Therefore, by the Intermediate Value Theorem, the functions f(x) and g(x) intersect at least twice between x = -2 and x = 4.