Given f'(x) = e^kx find k by finding the gradient of a tangent that intersects f(x) at (1,e^2) and also passes through the origin

Question

A curve for which f'(x)= e^kx where k is a constant, is such that the tangent at (1,e^2) passes through the origin find the gradient of the tangent to determine k and then integrate to find f(x)

how is it possible to find k by finding the gradient of the tangent that passes through f(x), i know that you can solve for k by subbing in (1,e^2) however that is just chance as it is not common for the tangent, f(x) and f'(x) to all intersect at the same point. so how can i solve for k by just finding the gradient of the tangent

The answers are k = 2 and f(x) = 1/2e^2x + 1/2e^2

Answer 1

Substitute x=1 in the expression for f'(x): f'(1)=e^k=e², i.e. k=2.

f'(x)=e^2x and f(x)=(1/2)e^2x + C

The equation of the tangent line is e²=y/x, i.e. y=xe² and at x=1, the value of the tangent is e²

The tangent and the original curve must both go through (1,e²)

Therefore (1/2)e² + C = e² and C = (1/2)e²

which gives f(x)=(1/2)e^2x + (1/2)e²

Answer 2

If I understand the problem, we first want to find the slope of the tangent line which f'(x) tells you, So

f'(x) = ek , therefore, the equation of the tangent line is y-e² = e^k(x-1) or y = xe^k +(e²-e^k)

So you know that the slope of the perpendicular line to this line is -1/e^k

So, the perpendicular line equation is y_p - e² = -(1/e^k)(x-1) or y_p = -(1/e^k)(x-1) + e²

Now, in order for this line to pass through the origin, we set (x,y) to (o,o) to get 0 = (1/e^k) + e²

Solving for k, we find that k = 2 and the perpendicular line is y_p = -(1/e²)(x-1) + e²

If f'(x) = e^kx = e^2x, then f(x) = (1/2)e^2x + constant to find the constant we substitute

f(1)= e² = (1/2)e² + constant or constant = (1/2)e²

Then k = 2 and f(x) = (1/2)e^2x + (1/2)e²