Given f"(x)=2x-6and f '(-3)=-4 and f(-3)=-1 find f '(x)and f(3)?

Question

Given f"(x)=2x-6and f '(-3)=-4 and f(-3)=-1 find f '(x)and f(3)

Answer 1

Let f"(x) = 2x - 6. Then,
f'(x) = integral ( f''(x) ) dx
          = integral ( 2x – 6 ) dx
          = 2 * (1/2)*x^2 – 6x + C
          = x^2 – 6x + C.

Thus,
f'(–3) = (–3)^2 – 6(–3) + C
          = 9 + 18 + C
          = 27 + C.

But f'(–3) = –4 as given above. So we can solve for C as follows
f'(–3) = –4 = 27 + C.

This tells us that C = –4 – 27 = –31.

Therefore, f'(x) = x^2 – 6x – 31.

Next, we determine f(x). That is,
f(x) = integral ( f'(x) ) dx
          = integral ( f'(x) ) dx
          = integral (x^2 – 6x – 31) dx
          = (1/3)*x^3 – 6*(1/2)*x^2 – 31x + K.
          = (1/3)x^3 – 3x^2 – 31x + K.

So we have,
f(–3) = (1/3)*( –3)^3 – 3(–3)^2 – 31(–3) + K
            = –9 – 27 + 93 + K
            = 57 + K.

Since f(–3)= –1, we equate this to the previous equation to obtain K, that is:
f(3) = –1 = 57 + K.

This implies that K = – 58. Thus,
f(x) = (1/3)*x^3 – 3x^2 – 31x – 58.

Moreover, we have
f(3) = (1/3)*(3)^3 – 3(3)^2 – 31(3) – 58
          = 9 – 27 – 93 – 58
          = – 169.

Answer 2

Given f"(x)=2x-6 ; f'(-3)=-4 ; f(-3)=-1
Find f '(x)and f(3)

f''(x) = 2x - 6
f'(x) = x² - 6x + C
f(x) = x³/3 - 3x² + Cx + D

f'(-3) = x² - 6x + C
-4 = (-3)² - 6(-3) + C
-4 = 9 + 18 + C
C = -31 so
f'(x) = x² - 6x - 31 and
f(x) = x³/3 - 3x² - 31x + D

f(-3) = x³/3 - 3x² - 31x + D
-1 = (-3)³/3 - 3(-3)² - 31(-3) + D
-1 = -9 - 27 + 93 + D
D = -58

f(-3) = x³/3 - 3x² - 31x - 58
f(3) = x³/3 -3x² - 31x - 58
f(3) = 9 - 27 - 93 - 58
f(3) = -169

So f'(x) = x² - 6x - 31 and f(3) = -169

Answer 3

f''(x) = 2x - 6
Integrate to get : f'(x) = x^2 - 6x + c1

f'(-3) = (-3)^2 - 6(-3) + c1 = -4
Therefore, c1 = -4 - 9 - 18 = -31

Thus, f'(x) = x^2 - 6x - 31

Integrate to get : f(x) = x^3 / 3 - 3x^2 - 31x + c2
f(-3) = (-3)^3 / 3 - 3(-3)^2 - 31(-3) + c2 = -1
Therefore, c2 = - 1 + 9 + 27 - 93 = -58

Thus, f(x) = x^3 / 3 - 3x^2 - 31x - 58

f(3) = 3^3 / 3 - 3(3)^2 - 31(3) - 58
= 9 - 27 - 93 - 58
= -169

Answer 4

A,B : constant

f"(x)=2x-6
　⇒　f '(x)=x^2-6x+A

f '(-3)=-4
　⇒　f '(-3)=(-3)^2-6*(-3)+A=-4
　　　A=-31

　∴　f '(x)=x^2-6x-31

f '(x)=x^2-6x-31　
　⇒　f (x)=(1/3)x^3-3x^2-31x+B

f (-3)=-1　⇒
f (-3)
=(1/3)*(-3)^3-3(-3)^2
-31*(-3)+B=-1
　　　B=-58

　∴　f (x)=(1/3)x^3-3x^2-31x-58

.f (x)=(1/3)x^3-3x^2-31x-58　⇒
f (3)
=(1/3)*(3)^3-3(3)^2-31*(3)-58

　∴　f (3)=-169

Answer 5

from the condition you offered,we come to know that
the f(-3) has two answers ,one is -4,the other is -1
.it is contraditory.paradox is the word i want to use to describe your statement