The given curve is y=x3-x+1.
=\(\frac{dy}{dx}\)=3x2-1
The slope of the tangent to a curve at \((x_0, y_0)\) is \((\frac{dy}{dx})\bigg] _{(x_0,y_0)}\).
It is given that x0 = 2.
Hence, the slope of the tangent at the point where the x-coordinate is 2 is given by,
\((\frac{dy}{dx}) \bigg]_{x=2}\)=3x2-1]x=2=3(2)2-1=12-1=11.
If f (x) = 3x2+15x+5, then the approximate value of f (3.02) is
It is given that at x = 1, the function x4−62x2+ax+9 attains its maximum value, on the interval [0, 2]. Find the value of a.
Find the maximum profit that a company can make, if the profit function is given by p(x) = 41−24x−18x2
What is the Planning Process?
m×n = -1