## Calculus 10th Edition

y=$\frac{79}{32}$ $\approx$ 2.47 x=1
(-0.5,y) y=$x^{5}$-5x Substitute x=-0.5 in the above equation to get the value of y at this point. y=$(-0.5)^{5}$-5 (-0.5) = $\frac{79}{32}$ $\approx$ 2.47 (x,-4) y=$x^{5}$-5x Substitute y=-4 in the above equation. You should get $x^{5}$-5x+4=0 By trial and error x=1 is a solution of the above equation $1^{5}$-5(1)+4=0 0=0