Answer:
The x-coordinates for which the gradient of the curve is 5 is at x = -1/2 and x = 1/2.
Step-by-step explanation:
We have the function:
[tex]\displaystyle y=\frac{x^2 -1}{x}[/tex]
And we want to find the x-coordinates for which the gradient (or slope) at the point is 5.
The gradients of a curve at a point is given by its derivative. Find the derivative:
[tex]\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left[\frac{x^2-1}{x}\right][/tex]
Use the quotient rule:
[tex]\displaystyle \frac{dy}{dx}=\frac{(x^2-1)'(x)-(x^2-1)(x)'}{(x)^2}[/tex]
Simplify:
[tex]\displaystyle \frac{dy}{dx}=\frac{(2x)(x)-(x^2-1)(1)}{x^2}[/tex]
Simplify:
[tex]\displaystyle \frac{dy}{dx}=\frac{x^2+1}{x^2}[/tex]
Since the gradient is 5, substitute 5 for dy/dx and solve for x:
[tex]\displaystyle 5=\frac{x^2+1}{x^2}[/tex]
Multiply both sides by x² (x ≠ 0):
[tex]5x^2=x^2+1[/tex]
Subtract x² from boths sides:
[tex]4x^2=1[/tex]
Solve for x:
[tex]\displaystyle x=\pm\sqrt{\frac{1}{4}}=\frac{1}{2}\text{ or } -\frac{1}{2}[/tex]
So, the x-coordinates for which the gradient of the curve is 5 is at x = -1/2 and x = 1/2.
