Question

Determine the location and values of the absolute maximum and absolute

minimum for the given function:

() = (− + 2)
^4

, ℎ 0 ≤ ≤ 3​

Answer 1

SOLUTION

TO DETERMINE

The location and values of the absolute maximum and absolute minimum for the given function:

$\sf{f(x) = {( - x + 2)}^{4} \: \: \: \: where \: \: 0 \leqslant x \leqslant 3 }$

EVALUATION

Here the given function is

$\sf{f(x) = {( - x + 2)}^{4} \: \: \: \: where \: \: 0 \leqslant x \leqslant 3 }$

Now

$\sf{f(x) = {(2 - x )}^{4} }$

Differentiating both sides with respect to x we get

$\sf{f'(x) = - 4{(2 - x )}^{3} }$

Now for critical points we have

$\sf{f'(x) = 0 }$

$\sf{ \implies \: - 4{(2 - x )}^{3} = 0}$

$\sf{ \implies \: {(2 - x )}^{3} = 0}$

$\sf{ \implies \: x = 2}$

Now we find the value of f(x) at the critical point x = 2 and at the extremities of the given interval i.e 0 and 3

$\sf{f(0) = {(2 - 0 )}^{4} = {2}^{4} = 16}$

$\sf{f(2) = {(2 - 2 )}^{4} = {0}^{4} =0}$

$\sf{f(3) = {(2 - 3 )}^{4} = {( - 1)}^{4} = 1}$

Thus we see that

∴ Absolute maximum value = 16 which occurs at x = 0

∴ Absolute minimum value = 0 which occurs at x = 2

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