Question

For the following function, find all its critical point(s) and its absolute extrema.
f(x)=10sqrt(x^2+10)−3x , 0<=x<=14

Answer 1

To find the critical point(s), we will look at the derivative of f and solve f'(x)=0:

f'(x) = 10x/[sqrt(x^2+10)] - 3 = 0

Make the left hand side into a single fraction and for the fraction to be zero, its numerator must be zero. So

10x - 3sqrt(x^2+10)=0
10x = 3sqrt(x^2+10)
100x^2 = 9(x^2 + 10)
x^2 = 90/91
x1 = sqrt(90/91) or x2 = -sqrt(90/91)

The critical points are (x1,f(x1)) and (x2,f(x2)). To find the absolute extrema, you can either use Second Derivative Test or First Derivative Test. The second derivative of f is

f"(x) = (x^3 -10x^2 +10x)/(x^2+10)^(3/2)

Answer 2

Some correction. Since x is between 0 and 14, only x1 above will be considered for the critical point. Besides that, since f'(x) is

f'(x) = [10x - 3sqrt(x^2+10)] / sqrt(x^2+10),

the critical point also can occur when the denominator, sqrt(x^2+10) of f' is 0. This comes from the definition of a critical number: a number c is a critical number of f if f'(c) = 0 or f'(c) does not exist.
But sqrt(x^2+10) = 0 has no solution, so only x1 above is the critical number for f.