If g(1) = 4 and g¢ (1) = − 2, find f ¢ (1), where f(x) = x2 ln (g(x)). Also find the equation of the
normal to f at x = 1.
I guess f ¢ means f'. So,
f(x) = x^2 ln(g(x))
f' = 2x ln(g(x)) + x^2/g(x) g'
f'(1) = 2 ln4 + -2/4 = (4ln4 - 1)/2
f(1) = ln4
The line normal to f at x=1 has slope 2/(1 - 4ln4)
So, the line through (1,4ln4) with that slope is
y - 4ln4 = 2/(1 - 4ln4) * (x-1)
That would be
So, the line through (1,ln4) with that slope is
y - ln4 = 2/(1 - 4ln4) * (x-1)
To find f ¢ (1), we first need to find the derivative of f(x) with respect to x. Let's break it down step by step.
Given f(x) = x^2 ln (g(x))
Step 1: Find the derivative of the first term, x^2:
f'(x) = 2x ln (g(x))
Step 2: Apply the product rule to the second term, ln(g(x)):
f'(x) = 2x ln (g(x)) + x^2 * [1/g(x)] * g'(x)
Step 3: Substitute x = 1 into the equation:
f'(1) = 2(1) ln (g(1)) + (1)^2 * [1/g(1)] * g'(1)
We are given g(1) = 4, so we can substitute that value:
f'(1) = 2(1) ln (4) + (1)^2 * [1/4] * g'(1)
Now, we also need to find g'(1). We are given g'(1) = -2, so we can substitute that value as well:
f'(1) = 2(1) ln (4) + (1)^2 * [1/4] * (-2)
Calculating this expression gives us f'(1) = 2 ln (4) - 1.
Now let's move on to finding the equation of the normal to f at x = 1.
The equation of the normal to a curve at a given point is of the form y - y1 = m(x - x1), where (x1, y1) is the point of tangency and m is the slope of the tangent line.
At x = 1, the point on the curve f(x) is (1, f(1)). From above, we know that f(1) = x^2 ln (g(1)), so we substitute x = 1 into the equation:
f(1) = (1)^2 ln (4)
f(1) = ln (4)
So the point of tangency is (1, ln (4)).
The slope of the tangent line is f'(1), which we already calculated as 2 ln (4) - 1.
Now let's substitute the values into the equation of the normal:
y - ln (4) = (2 ln (4) - 1)(x - 1)
This is the equation of the normal to f at x = 1.