Had a mock today and was asked to show that X=ln2 for the equation Y=(ex-2)2-1 And then asked to find the Y value. How would this be solved?
Some sort of incomplete question if that's all you're given. x=ln(2) is the x coordinate of the stationary point though so maybe something to do with that then you can plug it through the original to find the y coordinate and get a set of coordinates for the stationary point.
Had a mock today and was asked to show that X=ln2 for the equation Y=(ex-2)2-1 And then asked to find the Y value. How would this be solved?
If you're taking Calc I, maybe your teacher asked you to find the critical value of Y? Taking the derivative and setting it equal to 0 would give you X=ln2, studying the sign of Y′ on either side, and then you'd evaluate Y(ln2) to find the extrema.
If you're taking Calc I, maybe your teacher asked you to find the critical value of Y? Taking the derivative and setting it equal to 0 would give you X=ln2, studying the sign of Y′ on either side, and then you'd evaluate Y(ln2) to find the extrema.
You wouldn't need to do all this, of course. You've got a square plus some constant, so the function is minimized when the square is 0.