Give the range of P(x)

P(x) = √(6+5x-x^2)
Answer #1
4
Answer #2
12
Answer #3
http://www.wolframalpha.com/input/?i=P%28x%29+%3D+%E2%88%9A%286%2B5x-x^2%29
Answer #4
f(P,x)=~((6+5x-x^(2)))
The derivative of f(P,x) is equal to P'(x)=(d)/(dx) ~((6+5x-x^(2))).
P'(x)=(d)/(dx) ~((6+5x-x^(2)))
Find the derivative of the expression.
(d)/(dx) ~((6+5x-x^(2)))
Find the derivative of the expression.
(d)/(dx) ((6+5x-x^(2)))^((1)/(2))
Reorder the polynomial 6+5x-x^(2) alphabetically from left to right, starting with the highest order term.
((-x^(2)+5x+6))^((1)/(2))
Remove the parentheses around the expression -x^(2)+5x+6.
(-x^(2)+5x+6)^((1)/(2))
The chain rule states that the derivative of a composite function (f o g)’ is equal to (f’ o g)*g’. To find the derivative of (-x^(2)+5x+6)^((1)/(2)), find the derivatives of each portion of the function and use the chain rule formula.
(d)/(du) u^((1)/(2))*(d)/(dx) -x^(2)+5x+6
To find the derivative of u^((1)/(2)), multiply the base (u) by the exponent ((1)/(2)), then subtract 1 from the exponent.
(d)/(du) u^((1)/(2))=(1)/(2u^((1)/(2)))
To find the derivative of -x^(2), multiply the base (x) by the exponent (2), then subtract 1 from the exponent.
(d)/(dx) -x^(2)+5x+6=-2x+(d)/(dx) 5x+6
To find the derivative of 5x, multiply the base (x) by the exponent (1), then subtract 1 from the exponent (1-1=0). Since the exponent is now 0, x is eliminated from the term.
(d)/(dx) -x^(2)+5x+6=-2x+5+(d)/(dx) 6
Since 6 does not contain x, the derivative of 6 is 0.
(d)/(dx) -x^(2)+5x+6=-2x+5+0
Add 0 to 5 to get 5.
(d)/(dx) -x^(2)+5x+6=-2x+5
Replace the variable u with -x^(2)+5x+6 in the expression.
(d)/(du) u^((1)/(2))=(1)/(2(-x^(2)+5x+6)^((1)/(2)))
Replace the variable u with -x^(2)+5x+6 in the expression.
-2x+5
Form the derivative by substituting the values for each portion into the chain rule formula.
=(1)/(2(-x^(2)+5x+6)^((1)/(2)))*(-2x+5)
Multiply the rational expressions to get ((-2x+5))/(2(-x^(2)+5x+6)^((1)/(2))).
(d)/(dx) (-x^(2)+5x+6)^((1)/(2))=(-2x+5)/(2(-x^(2)+5x+6)^((1)/(2)))
The derivative of the function is ((-2x+5))/(2(-x^(2)+5x+6)^((1)/(2))).
P'(x)=(-2x+5)/(2(-x^(2)+5x+6)^((1)/(2)))
Answer #5
^There’s an easier way than this one, although some mathematicians will argue that it’s not the proper way to do it.
P(x) = sqrt(6+5x-x^2) >=0
P^2(x)=-x^2+5x+6
with maximum at x=-b/2a=5/2 which is P^2(5/2)=49/4
thus maximum of P^2(x)<=49/4 which means that P(x)<=7/2
The minimum for P^2(x) is at -oo, but P(x)>=0. So the minimum lies where P^2(x)=0 (at -1 and 6) and is equal to 0.