Question

What are the critical values of `f(x)=e^(x^2)-x^2+xe^x-5x+6e^x`?

Answer 1

`x approx -0.285`

Explanation:

Critical values are when the slope is zero. We can immediately recognize that this function has `x` in both the exponent and as coefficients, hence we will not be able to analytically solve this as its derivative will still be transcendental. We will, therefore, give numerical answers.

Taking the derivative of `f`,
`f(x) = e^(x^2) - x^2 + xe^x - 5x + 6e^x`
`f'(x) = 2x e^(x^2) - 2x + xe^x + e^x - 5 + 6e^x`

Plotting that,
graph{2x e^(x^2) - 2x + xe^x + e^x - 5 + 6e^x [-3, 3, -30, 30]}

we see that the curve has only a single zero (this is kinda obvious given that `e^(x^2)` grows much much faster than any other function and therefore dominates very quickly).

This zero is our one critical value at `x approx -0.285`.