Use calculus to find the absolute minimum of f(x) = x2 – 6x + 2

Question

1. Use calculus to find the absolute minimum of f(x) = x² – 6x + 2

2. Where is f(x) = -sin x + cos x concave up on [0, 2π] (it may help to consider graphs)

3. Evaluate two different ways: lim (x² – 16) / (x -4)
                        

                                                       X →4

Answer 1

1) derivative of f(x) = 2x -6

set = to zero, x = 3

find f(3) = 9 -18 +2 = -7 = absolute & relative minimum

that's using differential calculus

or use algebra and complete the square

x^2 -6x +2

= (x^2 -3)^2 +2-9

= (x^2-3)^2 -7 is in vertex form where (h,k) = (3,-7) = minimum point on the parabola

2) f(x) = -sinx + cosx

f'(x)= -cosx -sinx =0

sinx =-cosx

x = pi/4 = reference angle, but in quadrant 2 or 4 where sines and cosines have opposite signs

so

x= 3pi/4 or 7pi/4

now plug them into f(x) and see which is largest

f(3pi/4) = -sin(3pi/4) + cos(3pi/4)= 0

f(7pi/4) = -sin(7pi/4) + cos(7pi/4) = sqr2/2 + sqr2/2 = sqr2 = about 1.41421 = absolute and relative maximum

or use a graphing calculator like desmos to find or check your answer

3) (x^2-16)/(x-4) = x+4 graphically a straight line, but with a "hole" in it, a discontinuity at the point (4,8) limit = 8 = the y coordinate when x = 4

but plug in x=4 into f(x) you get f(4) = 0/0 which is undefined. so use L'Hopital's rule

take derivatives separately of numerator and denominator to get 2x and 1, now plug in x=4 to get 8