Find all critical numbers:
2x*sqrt(3x^2+4)?
is this
y = 2x*sqrt(3x^2+4) ?
critical number are
x and y intercepts, max/min points and points of inflection
for x-intercept, let y = 0
for y-intercept , let x = 0
for max/min , take first derivative, set equal to zero and solve for x. Sub back into original to get the matching y
for pts. of inflection, set second derivative to zero and solve for x
sub back into original to get the y
use product rule to find first derivative.
To find the critical numbers of the function f(x) = 2x√(3x^2+4), we need to take the derivative of the function and set it equal to zero.
Firstly, let's take the derivative of f(x) using the product rule and chain rule:
f'(x) = 2√(3x^2 + 4) + 2x * (1/2) *(3x^2 + 4)^(-1/2) * 6x
Simplifying further:
f'(x) = 2√(3x^2 + 4) + 6x^2 / √(3x^2 + 4)
Now we need to set this derivative equal to zero and solve for x:
2√(3x^2 + 4) + 6x^2 / √(3x^2 + 4) = 0
To make things easier, let's multiply every term by the common denominator √(3x^2 + 4):
2(3x^2 + 4) + 6x^2 = 0
Expanding and simplifying:
6x^2 + 8 + 6x^2 = 0
12x^2 + 8 = 0
12x^2 = -8
x^2 = -8/12
x^2 = -2/3
Since the square of a real number cannot be negative, there are no real solutions to this equation. Therefore, there are no critical numbers for the function f(x) = 2x√(3x^2+4).