# Thread: Showing gradient of a curve (Q10)

1. ## Showing gradient of a curve (Q10)

A curve is such that $\displaystyle \frac{d^2y}{dx^2}=5x-4$. The gradient of the curve at the point $\displaystyle (-2,4)$ is $\displaystyle 3$.

Show that the gradient is never less than $\displaystyle -\frac{83}{5}$

2. Originally Posted by Punch
A curve is such that $\displaystyle \frac{d^2y}{dx^2}=5x-4$. The gradient of the curve at the point $\displaystyle (-2,4)$ is $\displaystyle 3$.

Show that the gradient is never less than $\displaystyle -\frac{83}{5}$
The gradient of the curve is$\displaystyle \frac{dy}{dx}=\int (5x-4)dx=\frac{5x^2}{2}-4x+c$

At (-2,4), dy/dx=3 so c= -15

$\displaystyle dy/dx=\frac{5x^2}{2}-4x-15$

Complete the square and you will notice that the minimum value of the gradient is -83/5.

3. $\displaystyle \frac{dy}{dx}=\frac{5x^2}{2}-4x-15$
$\displaystyle =\frac{5}{2}x^2-4x-15+(\frac{4}{2})^2-(\frac{4}{2})^2$
$\displaystyle =(\frac{5}{2}x-2)^2-15-(\frac{4}{2})^2$
$\displaystyle =(\frac{5}{2}x-2)^2-19$

minimum value=-19 which doesnt tally with the question of $\displaystyle -\frac{83}{5}$