Maxima and Minima

Question

Let f(x) = e^{2x-4x^2-7}

The derivative of f is f'(x) = _________

An equation for the tangent line to the curve y = f(x) at x = 0 is y = ____________

Answer 1

Here, f(x) = e^{(2x − 4x^2 − 7)}.

Also, f'(x) = d(2x − 4x² − 7)/dx times e^{(2x − 4x^2 − 7)} or (2 − 8x)e^{(2x − 4x^2 − 7)}.

The equation of the line tangent to the curve of f(x) = e^{(2x − 4x^2 − 7)} at x = 0

is given according to the slope-intercept form or y = mx + b. Then build e^{(2x − 4x^2 − 7)} = (2 − 8x)e^{(2x − 4x^2 − 7)}x + b

which (for x = 0) goes to e^-7 = 2e^-7(0) + b.

This forces b to e^-7 and the tangent line is written as y = (2e^-7)x + e^-7.