fourth degree polynomial p(x) whose graph is symmetric about the y-axis, and which has a y-intercept of 10, and global maxima at (1,13) and (-1, 13).
Why do we end up with f(x) instead of f(t) on the Second Fundamental Theorem of Calculus?
5x^2
To find a fourth-degree polynomial that satisfies the given conditions, we can follow these steps:
1. Since the graph is symmetric about the y-axis, the polynomial will only contain even powers of x.
2. Let's assume the form of the polynomial as follows: p(x) = ax^4 + bx^2 + c.
3. Since the y-intercept is 10, we know that p(0) = 10.
Substituting x = 0 into the polynomial, we get:
p(0) = a(0)^4 + b(0)^2 + c = 0 + 0 + c = c
So, c = 10.
Therefore, the equation for our polynomial now becomes: p(x) = ax^4 + bx^2 + 10.
4. Now, let's use the given information about global maxima to find the values of a and b.
At x = 1, the polynomial has a maximum of 13. Substituting x = 1 into the polynomial, we get:
p(1) = a(1)^4 + b(1)^2 + 10 = a + b + 10 = 13.
Similarly, at x = -1, the polynomial also has a maximum of 13. Substituting x = -1 into the polynomial, we get:
p(-1) = a(-1)^4 + b(-1)^2 + 10 = a + b + 10 = 13.
Now we have a system of two equations:
a + b + 10 = 13 (equation 1)
a + b + 10 = 13 (equation 2)
Subtracting equation 2 from equation 1, we get:
a + b + 10 - (a + b + 10) = 13 - 13
0 = 0
This means that both equations are satisfied regardless of the values of a and b.
5. Since we have two variables (a and b) and only one equation (either equation 1 or 2), we can choose any value for a or b.
Let's assume a = 1.
Then, using equation 1:
1 + b + 10 = 13
b + 11 = 13
b = 13 - 11
b = 2
Therefore, our polynomial is: p(x) = x^4 + 2x^2 + 10.
To summarize, the fourth-degree polynomial that satisfies the given conditions is p(x) = x^4 + 2x^2 + 10.
If it is even, P(x)=ax^4+bx^2+c
P(0)=10 so c is 10
p'=5ax^3+2bx=0
x(5ax^2+2b)=0
maximum at x=+-1
so 5ax^2=-2b
or 5a=-2b or a=-5/2 b
but P(1)=13=(-5/2 b)*1+ b(1)+10
or 3/2 b=10 b=20/3 and a=-5/2 * 20/3=100/6
check P(-1)=13
it checks.