Question

Help me solve this problem!!!

Let p(x) = 3^x. A general quadratic polynomial can be written as q(x) = ax^2+ bx + C.
Find the quadratic polynomial q(x) such that:
q(0)=p(0)
q’(0)=p'(0)
q'(0)=p"(0).
q(X)
=

Answer 1

q(x) ≈ 0.60347x^2 +1.09861x +1

Explanation:

The derivative of an exponential function is ...

d(a^x)/dx = ln(a)·a^x

Then the second derivative is ...

d²(a^x)/dx² = ln(a)²·a^x

Here, you have a=3, so ...

q(0) = 3^0 = 1

q'(0) = ln(3)·3^0 = ln(3) ≈ 1.09861

q''(0) = ln(3)²·3^0 = ln(3)² ≈ 1.20695

__

The derivatives of p(x) are ...

p'(x) = 2ax +b ⇒ p'(0) = b = q'(0)

p''(x) = 2a ⇒ a = q''(0)/2

So, ...

q(x) = 1.20695/2x^2 +1.09861x +1

q(x) ≈ 0.60347x^2 +1.09861x +1