How to find range in example 21? Share with your friends Share 0 Atul Shaw answered this Dear student,f(x)=x2+3x+5x2-5x+4x2+3x+5D=b2-4ac=32-4×5×1=9-20=-11As. D<0 and (coeffiecient of x2)>0So, the value of x2+3x+5 >0x2-5x+4≠0x2-4x-x+4≠0x(x-4)-1(x-4)≠0(x-1)(x-4)≠0So x≠1 and x≠4Therefore, Df∈(-∞,∞)-{1,4}The range can't be easily calculated for such curve, we need to trace the graph and then draw the curve and then find the range.at x→∞ f(x)=limx→∞x2+3x+5x2-5x+4=limx→∞1+3x+5x21-5x+4x2=1at x→-∞ f(x)=limx→-∞x2+3x+5x2-5x+4=limx→-∞1+3x+5x21-5x+4x2=1at x→1- f(x)=limx→1-x2+3x+5x2-5x+4=limx→1-1+3×1+5(x-4)(x-1)=limx→1-1+3×1+5(1--4)(1--1)=∞+at x→1+ f(x)=limx→1+x2+3x+5x2-5x+4=limx→1+1+3+5(x-4)(x-1)=limx→1+1+3×1+5(1+-4)(1+-1)=∞-at x→4- f(x)=limx→4-x2+3x+5x2-5x+4=limx→4-1+3×1+5(x-4)(x-1)=limx→4-1+3×1+5(4--4)(4--1)=∞-at x→4+ f(x)=limx→4+x2+3x+5x2-5x+4=limx→4+1+3+5(x-4)(x-1)=limx→4+1+3×1+5(4+-4)(4+-1)=∞+f'(x)=x2-5x+4(2x+3)-x2+3x+5(2x-5)x2-5x+42=0x2-5x+4(2x+3)-x2+3x+5(2x-5)=02x3-10x2+8x+3x2-15x+12-2x3-6x2-10x+5x2+15x+25=0-8x2-2x+37=08x2+2x-37=0x=2.03 and x=-2.28The differentiation gives us a local maxima or minima, so, here we are checking for x∈(1,4)so, x=2.03f(2.03)=x2+3x+5x2-5x+4=2.032+3×2.03+52.032-5×2.03+4=-7.5As, at x→1+ and x→4- we have f(x)→∞-, it means at x=2.03, it is a maximaso, Rf∈-∞,∞--7.5,1Curve tracing was too complex so it would give an approximate ans and not the exact oneRegards 0 View Full Answer
