Find the critical numbers of the function.?

Question

I cant find this

f(x)= xe ^ - 9x


and for this question

b) f(x)= x^3 +7x^2 - 24x

f'(x) = 3x^2 + 14x - 24

But what do I do next please help me fast!

Answer 1

1)
f(x) = xe^(-9x)
f'(x) = e^(-9x) + xe^(-9x)(-9)
f'(x) = e^(-9x) - 9x e^(-9x)
f'(x) = e^(-9x) [ 1 - 9x ]

Critical values are what make f'(x) = 0 or what makes f'(x) undefined.

0 = e^(-9x) [ 1 - 9x ]

Equate each factor to 0.
e^(-9x) = 0
1 - 9x = 0

The first equation has no solutions. For the second equation,
1 = 9x, so x = 1/9. Critical value is x = 1/9.

2) f(x) = x^3 + 7x^2 - 24x

Like you said,

f'(x) = 3x^2 + 14x - 24

Set f'(x) = 0.

0 = 3x^2 + 14x - 24

Solve for x.

0 = (3x - 4)(x + 6)

3x - 4 = 0 (i.e. x = 4/3)
x + 6 = 0 (i.e. x = -6)

The critical numbers are x = {4/3, -6}

Answer 2

f(x) = x e^(-9x)
f'(x) = 1.e^(-9x) + x.e^(-9x).(-9)
= e^(-9x) (1 - 9x)
= 0 <=> x = 1/9.
f(1/9) = 1/9 e^(-1) = 1/(9e).
So the critical point of this function is at (1/9, 1/(9e)).

f(x) = x^3 + 7x^2 - 24x
f'(x) = 3x^2 + 14x - 24
= (3x - 4)(x + 6)
= 0 <=> x = 4/3 or -6.
So the critical points are at (4/3, -464/27) and (-6, 180).

Answer 3

LOL set the derivative of the function equal to 0 and solve for all x values that satisfy the equation.
f'(x) = e^ -9x - 9xe^-10x = 0
e ^ x = 9
x = ln9

3x^2 + 14x - 24 = 0
quadratic equation
[-14 +/- (14^2 + 12*24)]/6 = x
x = -83, 78.333333333
good luck!