If maximum and minimum values of x²-5x+6 in [0,4] are 'a' and 'b', th

1.	If maximum and minimum values of x²-5x+6 in [0,4] are 'a' and 'b', then a²+b²= ?Please Help
Answer» 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, $\rm :\longmapsto\: {a}^{2} + {b}^{2}$ $\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = {(6)}^{2} + {\bigg( \dfrac{ - 1}{4} \bigg) }^{2}$ $\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 36 + \dfrac{1}{16}$ $\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \dfrac{576 + 1}{16}$ $\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \dfrac{577}{16}$