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A function f is defined for all real numbers and has the following three properties: f(1)=5 \quad f(3)=21 \quad f(a+b)-f(a)=k a b+2 b^{2}f(1)=5f(3)=21f(a+b)−f(a)=kab+2b 2 for all real numbers a and b where k is a fixed real number independent of a and b. (a) Use a=1 and b=2 to find k. (b) Find f^{\prime}(3)f ′(3) (c) Find f^{\prime}(x)f (x) for all real x.



Answer :

The solutions of given function;

(a) k = -12

(b) f'(3) = 0

(c) f'(x) = 0

Given info,

A function f is defined for all real numbers and has the following three properties:

[tex]f(1)=5 \quad f(3)=21 \quad f(a+b)-f(a)=k a b+2 b^{2}f(1)\\[/tex]

for all real numbers a and b where k is a fixed real number independent of a and b.

To find,

(a) Use a = 1 and b = 2 to find k.

[tex]f(a+b)-f(a)=k a b+2 b^{2}f(1)[/tex]

f(1 + 2) - f(1) = k(1 * 2) + 2 * 2² * f(1)

f(3) - f(1) = 2k + f(1) * 8

21 - 5 = 2k + 8 * 5

16 = 2k + 40

-2k = 24

k = -12

(b) Find f'(3).

f(3) = 21

f'(3) = d/dx(21) = 0

(c) Find f'(x) for all real x.

f'(x) = 0

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