Question

If maximum and minimum values of x²-5x+6 in [0,4] are 'a' and 'b', then a²+b²= ?

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Answer 1

Basic Concept used

Finding maxima and minina of f(x) on [a,b]

• Verify that the function is continuous on the interval [a,b] .

• Find all critical points of f(x) that are in the interval [a,b].

• Evaluate the function at the critical points found in step 1 and the end points.

• Identify the maxima and minima.

Let's solve the problem now!!.

$\rm :\longmapsto\:Let \: f(x) = {x}^{2} - 5x + 6$

Differentiate w. r. t. x, both sides we get

$\rm :\longmapsto\:f'(x) = 2x - 5$

For maxima or minima,

$\rm :\longmapsto\:f'(x) = 0$

$\rm :\longmapsto\:2x - 5 = 0$

$\rm :\implies\:x = \dfrac{5}{2}$

$\rm :\implies\: x = 0, \: \dfrac{5}{2} ,4$

So,

$\rm :\longmapsto\:f(0) = {0}^{2} - 5 \times 0 + 6 = 6$

$\rm :\longmapsto\:f\bigg(\dfrac{5}{2} \bigg) = {\bigg(\dfrac{5}{2} \bigg) }^{2} - 5 \times \dfrac{5}{2} + 6$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \dfrac{25}{4} - \dfrac{25}{2} + 6$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \dfrac{25 - 50 + 24}{4}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = - \: \dfrac{1}{4}$

$\rm :\longmapsto\:f(4) = {4}^{2} - 5 \times 4 + 6$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 16 - 20 + 6$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2$

$\rm :\implies\:Maximum \: value \: is \: = \: 6$

and

$\rm :\implies\:Minimum \: value \: = - \: \dfrac{1}{4}$

So,

It means,

$\rm :\longmapsto\:a = 6 \: \: and \: \: b = - \dfrac{1}{4}$

Hence,

$\rm :\longmapsto\: {a}^{2} + {b}^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = {(6)}^{2} + {\bigg( \dfrac{ - 1}{4} \bigg) }^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 36 + \dfrac{1}{16}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \dfrac{576 + 1}{16}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \dfrac{577}{16}$