Calculus Question

Question

Find the absolute maximum and absolute minimum values of f on the given interval.

f(x) = x

x2 − x + 25

,  [0, 15]

absolute minimum value
absolute maximum value

Answer 1

f(15) = 15^2 - 15 +25 = 225+10 = 235 = the absolute maximum in the given interval [0,15]

it's an upward opening parabola with vertex and absolute and local minimum at x=1/2, f(1/2) = 24.25

f(0) = 25

f(1/2) = (1/2)^2 -(1/2) + 25 = 1/4 + 24.5 = 24.25. (0.5, 24.25) is the vertex. axis of symmetry is x= 1/2

f'(x) = 2x -1= 0

2x= 1

x = 1/2 is the x value for the vertex

f(1/2) = 24 1/4 < f(0) = 25 < f(15) = 235

for f(x) = x

absolute maximum is f(15) = 15

absolute minimum is f(0) = 0