4.5

 Use zero- through third-order Taylor series expansions to predict f(3)
 for

 f(x) = 25*x^3 - 6*x^2 +7*x - 88

using a base point at x = 1

x = 0:0.01: 3.5;

f_x = @(x) 25*x.^3 - 6 * x.^2+7*x -88;

f1_x = @(x) 3* 25*x.^2 - 2* 6 * x.^1+7;

f2_x = @(x) 2* 3* 25*x - 2* 6;

f3_x = @(x) 2* 3* 25;

T_0_1 = f_x(1) * x.^0;

T_1_1 = T_0_1 + f1_x(1) * (x-1).^1;

T_2_1 = T_1_1 + f2_x(1) .* (x-1).^2/2;

T_3_1 = T_2_1 + f3_x(1) * (x-1).^3/6;

hold on

plot(x,f_x(x))             %blue

plot(x, T_0_1, 'r')       %red

plot(x, T_1_1, 'm')     %magenta

plot(x, T_2_1, 'y')      %yellow

%plot(x, T_3_1, 'g')   %green

hold off

With the third order Taylor plot hidden, the graph looks like 

Showing that with increasing orders of the taylor series, the curve more closely approximates the original equation, until we show the third order equation, at which point

one can see that the third order taylor exactly follows the original equation. This stands to reason, as it is mentioned in the text that in general, the nth order Taylor series will be exact for an nth order polynomial, and the equation we are approximating has a largest order term of 25x^3, making it 3rd order.