Let 𝑓(𝑥)=𝑥2−6𝑥+10. Find the critical point c of f(x) and compute f(c).

Question

Let 𝑓(𝑥)=𝑥²−6𝑥+10.

Find the critical point  c of  f(x) and compute f(c).

The critical point c is = 

The value of  f(c)= 

Compute the value of  at the endpoints of the interval [0,6].

 f(0)= 

 f(6)= 

Determine the min and max of f(x) on [0,6].

Minimum value = 

Maximum value = 

Find the extreme values of f(x) on [0,1].

Minimum value = 

Maximum value = 

Answer 1

To find critical point c, solve f'(c) = 0

f'(x) = 2x-6, so x = 3 when f'(x) = 0, so c = 3

f(c) = 9 - 18 + 10 = 1

f(0) = 10

f(6) = 10

f(x) is a quadratic, facing up, so it will have a mimimum at x=3, value f(3) = 1

To find the minimum on [0,6]. As there is a relative minimum at x=3, we find the minimum value of f(x) here and at the end points.

f(0) = 10

f(3) = 1

f(6) = 10

The minimum of these is 1, when x =3. So the minimum value on [0,6] is 1

Maximum.

No maximum between 0 and 6.

f(0)=10

f(6) = 10

So the maximum value is 10, at x=0 and x=6

Extreme values of f(x) on [0,1]

Minimum when x=1, will be 5

Maximum when x=0, will be 10