A function is given by, f(x)=e^-3x

write down the third-order Taylor approximation for f(x) about x =0.
Hence, evaluate f(0.258) giving your answer to 4 decimal places.

f = e^-3x

f' = -3e^-3x
f'' = 9e^-3x
f''' = -27e^-3x

f(x) = 1 - 3x + 9/2 x^2 - 9/2 x^3
f(.258) = 0.4483

e^(-3*.258) = 0.4612

i was going around in circles and putting the wrong values.

thank you for your help. thanks.

To find the third-order Taylor approximation for a given function, we need to calculate the function's value and its derivatives at the specified point.

The general formula for the third-order Taylor approximation for a function f(x) about x=a is:

T(x) = f(a) + f'(a)(x-a) + (f''(a)(x-a)^2)/2! + (f'''(a)(x-a)^3)/3!

Let's compute the derivatives of f(x)=e^-3x:

f'(x) = -3e^-3x (by applying the chain rule)
f''(x) = 9e^-3x (by applying the chain rule again)
f'''(x) = -27e^-3x (by applying the chain rule a third time)

Now we can substitute x=0 into each derivative to find the values at x=0:

f(0) = e^0 = 1
f'(0) = -3e^0 = -3
f''(0) = 9e^0 = 9
f'''(0) = -27e^0 = -27

Using these values, we can write the third-order Taylor approximation:

T(x) = f(0) + f'(0)(x-0) + (f''(0)(x-0)^2)/2! + (f'''(0)(x-0)^3)/3!

T(x) = 1 + -3x + (9x^2)/2 + (-27x^3)/6

Simplifying further:

T(x) = 1 - 3x + (9x^2)/2 - 9x^3/2

To evaluate f(0.258) using this approximation, we substitute x = 0.258 into T(x):

f(0.258) = 1 - 3(0.258) + (9(0.258)^2)/2 - 9(0.258)^3/2

=f(0.258) ≈ 0.9424 (rounded to 4 decimal places)