If f(x)=ax^4-bx^2+x+5 and f(-3)=2, then what is the value of f(3)?

Solution given:

f(x)=ax^4-bx^2+x+5 and

f(-3)=2

f(-3)=a(-3)^4-b(-3)^2+(-3)+5

2=81a-9b-3+5

2=81a-9b+2

subtracting both side by 2 and adding 9b

2-2+9b=81a-9b+9b+2-2

9b=81a

now

f(3)=a(3)^(4)-b(3)^(2)+3+5

f(3)=81a-9b+8

substituting value of 81a

f(3)=9b-9b+8

f(3)=8

answer: b.48 for

step-by-step explanation:

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step-by-step explanation: