HOW TO FIND THE SLOPE OF A CURVE AT A GIVEN POINT

Find the gradient of the tangent to

Problem 1 :

f(x) = 3x² at x = -1

Solution :

f(x) = 3x²

f'(x) = 3(2x)

= 6x

Apply x = -1 in the first derivative.

f'(-1) = 6(-1)

= -6

So, -6 is the slope of the tangent line for the curve drawn at the point x = -1

Problem 2 :

f(x) = 6/x at x = 2

Solution :

f(x) = 6/x

Here x is at the denominator, we can move the x to the numerator and find the derivative.

f(x) = 6x^-1

f'(x) = -6x^-2

f'(x) = -6/x²

Slope at x = 2

f'(2) = -6/2²

= -6/4

= -3/2

So, the required slope is -3/2.

Problem 3 :

f(x) = x² + 2x + 1 at x = 1

Solution :

f(x) = x² + 2x + 1

f'(x) = 2x + 2(1) + 0

= 2x + 2

Slope at x = 1

f'(1) = 2(1) + 2

= 2 + 2

f'(1) = 4

So, the required slope is 4.

Problem 4 :

f(x) = x² + 7x at x = -2

Solution :

f(x) = x² + 7x

f'(x) = 2x + 7(1)

= 2x + 7

Slope at x = -2

f'(-2) = 2(-2) + 7

= -4 + 7

f'(-2) = 3

So, the required slope is 3.

Problem 5 :

f(x) = (x² + 1)/x at x = 2

Solution :

f(x) = (x² + 1)/x at x = 2

Here the variable x is at both numerator and in the denominator, so we have to use quotient rule to find the derivative.

u = x² + 1 and v = x

u' = 2x + 0 ==> 2x, v' = 1

Applying the quotient rule,

d(u/v) = (vu' - uv')/v²

= [x(2x) - (x² + 1)(1)]/x²

= (2x² - x² - 1)/x²

f'(x) = (x² - 1)/x²

Slope at x = 2

f'(2) = (2² - 1)/2²

= (4 - 1) / 4

= 3/4

So, the required slope is 3/4.

Problem 6 :

f(x) = √x + (8/x) at x = 4

Solution :

f(x) = √x + (8/x)

f(x) = √x + (8x^-1)

f'(x) = (1/2√x) - (8/x²)

Slope at x = 4

f'(4) = (1/2√4) - (8/4²)

= 1/2(2) - (8/16)

= 1/4 - 1/2

= (1 - 2)/4

= -1/4

So, the required slope is -1/4.