Question

Let f(x)=ax^7+bx^3+cx-5, where a,b and c are constants.If f(-7)=7,find the value of f(7).

Answer 1

Answer:

-17

Step-by-step explanation:

f(-7) = -a(7)^7 - b(7)^3 - 7c - 5 = 7

f(7) = a(7)^7 + b(7)^3 + 7c - 5 = t

Adding both the equations,

t + 7 = -5 -5

t = -17

Hence, f(7) = -17

Answer 2

The value is f(7)=-17.

Step-by-step explanation:

Given : Function $f(x)=ax^7+bx^3+cx-5$  where a,b and c are constants. If f(-7)=7.

To find : The value of f(7) ?

Solution :

$f(x)=ax^7+bx^3+cx-5$

Substitute x=-7,

$f(-7)=a(-7)^7+b(-7)^3+c(-7)-5$

$-a(7)^7-b(7)^3-c(7)-5=7$  ....(1)

Substitute x=7,

$f(7)=a(7)^7+b(7)^3+c(7)-5$

$a(7)^7+b(7)^3+c(7)-5=x$  ....(2)

Add equation (1) and (2),

$-a(7)^7-b(7)^3-c(7)-5+a(7)^7+b(7)^3+c(7)-5=7+x$

$-10=7+x$

$x=-17$

Therefore, the value is f(7)=-17.

#Learn more

If f(x) =ax^ + bx+ c,f(1)= 3,f(2)=7,f(3)=13 . Find the values of a,b and f(0)

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