Answer:
(-2.4, 37.014)
Step-by-step explanation:
We are not told how to approach this problem.
One way would be to graph f(x) = x^5 − 10x^3 + 9x on [-3,3] and then to estimate the max and min of this function on this interval visually. A good graph done on a graphing calculator would be sufficient info for this estimation. My graph, on my TI83 calculator, shows that the relative minimum value of f(x) on this interval is between x=2 and x=3 and is approx. -37; the relative maximum value is between x= -3 and x = -2 and is approx. +37.
Thus, we choose Answer A as closest approx. values of the min and max points on [-3,3]. In Answer A, the max is at (-2.4, 37.014) and the min at (2.4, -37.014.
Optional: Another approach would be to use calculus: we'd differentiate f(x) = x^5 − 10x^3 + 9x, set the resulting derivative = to 0 and solve the resulting equation for x. There would be four x-values, which we'd call "critical values."
Answer:
A. Maximum point :(-2.4,37.014)
Minimum point: (2.4,-37.014).
Step-by-step explanation:
Given function
f(x)=[tex]x^5-10x^3+9x[/tex] on interval [tex]\left[ -3, 3\right][/tex]
A. Maximum point (-2.4,37.014)
Minimum point (2.4,-37.014)
f(x)=[tex](-2.4)^5-10(-2.4)^3+9(-2.4)[/tex]
f(x)=-79.62624+138.24-21.6
f(x)=37.014
Put x= 2.4 then we get
f(x)= [tex](2.4)^5-10(2.4)^3+9(2.4)[/tex]
f(x)=79.62624-138.24+21.6
f(x)=37.014
B. Maximum point (2.4,-37.014)
Minimum point (-2.4,37.014)
Put x= 2.4. Then we get
f(x)= -37.014
Put x=-2.4 then we get
f(x)= 37.014
C. Maximum point ( -1.4, 33.014)
Minimum point ( 1.4, -33.014)
Put x=-1.4 then we get
f(x)=[tex](-1.4)^5-10(-1.4)^3+9(1.4)[/tex]
f(x)=-38.94
Put x= 1.4 then we get
f(x)= 38.94
D. Maximum point ( -3,30)
Minimum point ( 3,-30)
Put x=3 then we get
f(x)= 243-270+27=0
Put x=-3 then we get
f(x)= -243+270-27=0
Hence , from option A,B,C and D we can see only option A is right answer.
The approximate values of the minimum point (2.4,-37.014) and maximum point (-2.4, 37.014) of the function f(x)=[tex]x^5-10x^3+9x[/tex] on [tex]\left[-3,3\right][/tex].
