Calculus question?

Question

For what value of c on [0,1] is the tangent to the graph of f(x) = e^x-x^2 parallel to the secant line?

Answer 1

f(x)=e^x-x^2
f'(x)=e^x-2x
on [0,1],
f'(0)=e^0-2(0)=1
so, the gradient of tangent=1

y=mx+c
m=1
on [0,1]
1=1(0)+c
c=1

So, the tangent line is
y=x+1

Answer 2

The secant line passes through (0,1) and (1,e-1)

Its slope is (e -2)

So we must find a point of f(x) where the derivative = e-2

e^x-2x =e-2 So x = 1 is a solution .There is another solution
between 0 and Ln2