Question

Consider the following function. F(t) = 315 – 2012 + 16 Find the derivative of the function. F"(t) = Find the critical numbers of the function. (Enter your answers as a comma-separated list.) t = Find the relative maxima and relative minima, if any, of the function. (If an answer does not exist, enter DNE.) relative maximum (x, y) = relative minimum (x, y) =

Answer 1

Selm: Flt) = 345 - 20 13 +16 A Ellie (158t - do t27 for critical points FlWzo o 1544-600fco + H4-4220 $? (+34)20 5 t=0, 2-4 co of=4 fui£ they critical points are at = 0, -212 . Now, we will calculate relative manina and relative manima or flacy)= e6x² + by a

My f(a,y)= 6x²+by? partial derivative of f(is) and by with respect to a andy New, fx = 127 e6 x² + 642 ty = 124 e 6x²+642 for critical points en and fy = 0 fa=0 layer2²+ry² : 122 e 6x² +6y2 12y = 0 >> 122= 0 2=0 syzo therefore critical points is (as) = (0,0) :: 422= 122. (2644644) + 262*484? (123) = 1228 122 € 62€7192 +26**+6yyje È & 14422 eri toy mahiyet try the 6x²try? fyy = 144 y² e 6x²+672 +1 266x?tsye fay = 144 ny 62²+ 6ye 0- 0 Now Da fox fyy - (fry 2 | - (144x + 3) + ( 44 *H13) at ligje (oro) IDR 12:12 - 0 E144.30 and fax (010)= oxe + 12 xe. - 04193.1270 Hence, flaiss has relative minimum at (0/0) thes relative minimum value of (010) = 6 6x0² +6xo e and relative manimum valve ONET oto eo relative minimum value o Ang