# Solve by using the Secant Method up to ex - 3x2 = 0 for 0 ≤ x ≤1 and 3 ≤ x ≤5 four iterations up2 four decimal places it is about numerical analysis

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### 1 Answer

We'll put the formula of secant method:

y = {[f(x1)-f(x0]/(x1-x0)}(x-x1) + f(x1)

We'll note f(x) = e^x- 3x^2 for 0 ≤ x ≤1 and 3 ≤ x ≤5

x0 = 0

x1 = 3

f(x1) = f(3) = e^3- 3*3^2

f(3) = e^3 - 27

f(x0) = f(0) = e0

f(0) = 1

We'll substitute f(x1) and f(x0) in the formula:

y = {[f(x1)-f(x0]/(x1-x0)}(x-x1) + f(x1)

y = (e^3 - 27 - 1)(x - 3)/(x - 0) + e^3- 27

y = (e^3 - 28)(x - 3)/x + e3 - 27

x = x1 - f(x1)(x1-x0)/[f(x1) - f(x0)]

x = 3 - (e^3 - 27)*3/2

x = (6 - 59.049 + 81)/2

**x = 13.9755**

x3 = x2 - f(x2)(x2-x1)/[f(x2) - f(x1)]

x2 = 5

f(5) = e^5- 3*25

f(5) = 68.4890

x3 = 5 - 68.4890*(5-3)/(68.4890 +7.317)

x3 = 5 + 136.978/75.806

**x3 = 6.8069**