Question

If f(x) = mx² - 6x - 3 and f'(1) = 12. find the value of constant in (a) 9 (b) 3 (c)-3 (d) -4​

Answer 1

Answer:

a) 186

b) 6

c) 42

Step-by-step explanation:

$f(x) = m {x}^{2} - 6x - 3$

f'(1) = 12

f'(x) = 2mx - 6

12 = 2(1)m - 6

12 - 6 = 2m

$\frac{2m}{2} = \frac{6}{2}$

$m = 3$

$f(x) = 2 {x}^{2} - 6x - 3$

a)

f(9) = 3( 9)^2 - 6(9) - 3

f(9) = 186

b)

f( 3) = 3( 3)^2 - 6( 3) - 3

f(3) = 6

c)

f( - 3) = 3( - 3)^2 - 6( - 3) - 3

f(-3) = 42