Question

At which point on the following curve does the tangent line has the largest slope ?

y=1+40x^3-3x^5

Answer 1

The slope of the tangent line to the function y at the point x is equal to the derivative of this function at the given point.

$y’=(1+40x^3-3x^5)’=120x^2-15x^4$

Now we should find the points at which $y’$ has maximum.

$y’’=240x-60x^3=0$

$x=-2$ ; $x=0$ ; $x=2$ .

$y’’(-3)=-720+27\cdot60>0$

$y’’(-1)=-240+60<0$

$y’’(1)=240-60>0$

$y(3)=720-27\cdot60<0$

We have two maximums $y’$ at the points $x=-2$ and $y=2$ .

$y’(-2)=y’(2)=120\cdot(-2)^2-15\cdot(-2)^4=240$

Answer: $(-2,240)$ and $(2,240)$ .