find the equation of a line tangent to h(x)=x^2g(x) at x=2

Question

find the equation of a line tangent to h(x)=x^2g(x) at x=2

Answer 1

f(2)=2^2=4

f'(x)= 2x

f'(2)= 2(2)=4 =slope of the rangent line

y-4= 4(x-2) is the tangent line

or

y=4x-4 = g(x)

g(x)= 4x-4 is the tangent line

Answer 2

The tangent line of h(x) at x = 2 will be given by y = h'(2)·(x - 2) + h(2)

By the product rule, h'(x) = 2x·g(x) + x²·g'(x), which means h'(2) = 4·g(2) + 4·g'(2).

Plugging in 2 into h(x) gives h(2) = 4·g(2).

Now plugging h'(2) and h(2) into the equation of tangent line, we get

y = (4·g(2) + 4·g'(2))·(x - 2) + 4·g(2)

y = 4x·g(2) + 4x·g'(2) - 8·g(2) - 8·g'(2) + 4·g(2)

y = 4(g(2) - g'(2))x - 4·g(2) - 8·g'(2)

Without knowing what the function g(x) is, we cannont proceed further, so we are done.