Find the concavity and min and max for f(x) = x*sqrt(x^2+36) defined on the interval [-4, 7]?
To find the concavity and min/max for the function f(x) = x*sqrt(x^2+36) on the interval [-4, 7], we need to find the critical points and the second derivative of the function.
First, let's find the critical points by solving f'(x) = 0. Differentiating f(x) using the product rule gives:
f'(x) = (d/dx) [x * sqrt(x^2 + 36)]
= sqrt(x^2 + 36) + x * (d/dx) [sqrt(x^2 + 36)]
= sqrt(x^2 + 36) + x * (1/2) * (x^2 + 36)^(-1/2) * 2x
= sqrt(x^2 + 36) + (x^2 / sqrt(x^2 + 36))
Setting f'(x) to zero and solving for x:
sqrt(x^2 + 36) + (x^2 / sqrt(x^2 + 36)) = 0
Squaring both sides:
x^2 + 36 + x^2 = 0
Simplifying:
2x^2 + 36 = 0
2x^2 = -36
x^2 = -18
Since the square root of a negative number is undefined in the real number system, there are no real solutions to this equation. Therefore, the function f(x) = x*sqrt(x^2+36) has no critical points on the interval [-4, 7].
Next, let's find the second derivative of f(x) by differentiating f'(x):
f''(x) = (d/dx) [sqrt(x^2 + 36) + (x^2 / sqrt(x^2 + 36))]
= (1/2) * (x^2 + 36)^(-1/2) * 2x + (d/dx) [(x^2 / sqrt(x^2 + 36))]
= x / sqrt(x^2 + 36) - (x^2 / sqrt(x^2 + 36)^3) * 2x
Simplifying f''(x):
f''(x) = x / sqrt(x^2 + 36) - (2x^3 / sqrt(x^2 + 36)^3)
Now, to determine the concavity, we need to find the sign of f''(x) on the interval [-4, 7]. To do this, we can evaluate f''(x) at any point within the interval and check its sign.
Let's evaluate f''(x) at the endpoints of the interval, x = -4 and x = 7:
f''(-4) = (-4) / sqrt((-4)^2 + 36) - (2(-4)^3 / sqrt((-4)^2 + 36)^3)
= -4 / sqrt(16 + 36) - (2(-64) / sqrt(16 + 36)^3)
= -4 / sqrt(52) - (-128 / sqrt(52)^3)
= -2 / sqrt(13) + (-4 / sqrt(13)^3)
= (-2sqrt(13) - 4) / sqrt(13)^3
f''(7) = (7) / sqrt((7)^2 + 36) - (2(7)^3 / sqrt((7)^2 + 36)^3)
= 7 / sqrt(49 + 36) - (2(343) / sqrt(49 + 36)^3)
= 7 / sqrt(85) - (686 / sqrt(85)^3)
= (7sqrt(85) - 686) / sqrt(85)^3
From the signs of f''(-4) and f''(7), we can conclude that:
- f''(-4) < 0, meaning the concavity is downward (concave down) at x = -4.
- f''(7) > 0, meaning the concavity is upward (concave up) at x = 7.
Since there are no critical points on the interval [-4, 7], we cannot find the minima or maxima of the function f(x) = x*sqrt(x^2+36) on this interval.