Slopes and Tangents Calculus

Question

a) Normal to a curve Find an equation for the line perpendicular to the tangent to the curve $y = x^3 - 4x + 1$ at the point $(2,1)$.

b)Smallest Slope What is the smallest slope on the curve?At what point on the curve does the curve have this slope?

c)Tangents having specified slope Find equations for the tangents to the curve at the points where the slope of the curve is $8$.

The first question is fairly obvious as differentiating the curve gives you the slope at the given point and then finding the equation of a line perpendicular to that tangent is fairly easy, I'm stuck with the next question where I'm asked to find the smallest slope on the curve.

Is there a property or any particular arithmetical way to determine the same?

Answer 1

For part (b): The slope at a point on the graph of the given equation is given by $y'=3x^2-4$ i.e. by $f(x)=3x^2-4$. The smallest slope is given by the minimum value of $f$. It's not difficult to see that $f(x)$ is a parabola so the minimum value is attained at the local minimum which occurs at $x=0$, namely at point $(0,1)$ on the curve you have. Then $$\min{f}=f(0)=-4.$$

EDIT: After you differentiate and find the function whose value is the slope e.g. $f(x)=3x^2-4$ you then need to find the minimum. Finding local minimum: $f'(x)=6x$, set this equal to $0$: $f'(x)=0$ when $x=0$. Then $f''(x)=6>0$ so the critical point $x=0$ of $f$ is a local minimum. Since $$ \lim_{x\rightarrow -\infty} f(x)= +\infty \quad \lim_{x\rightarrow +\infty} f(x)=+\infty $$ we deduce that the global minimum of $f(x)$ is the local minimum value which is $f(x)|_{x=0}=f(0)=-4$.