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India
Class XI

Tangents (gradients and equations using power rule)

Lesson

This means using the derivative we can 

  • find the gradient function, also called the derivative
  • evaluate the gradient at a point, say $x=a$x=a
  • find the equation of the tangent at that point, because we have the gradient and a point

 

Let's work through an example

Find the equation of the tangent to the curve $f(x)=\frac{1}{4}x^2-2x+3$f(x)=14x22x+3 at the point where $x=4$x=4

Think:

This is our plan of attack

a) find the derivative function $f'(x)$f(x)

b) find the value of the derivative at the point where $x=4$x=4, this will be the gradient of the tangent

c) find the full coordinate on the curve by finding $f(4)$f(4)

d) use the point and the gradient to find the equation of the tangent.

 

a) find the derivative function $f'(x)$f(x)

The function is $f(x)=\frac{1}{4}x^2-2x+3$f(x)=14x22x+3, so using the power rule on each term we find that the derivative is $f'(x)=\frac{x}{2}-2$f(x)=x22

b) find the value of the derivative at the point where $x=4$x=4, this will be the gradient of the tangent.

$f'(x)=\frac{x}{2}-2$f(x)=x22, so $f'(4)=\frac{4}{2}-2=0$f(4)=422=0

This means that at the point where x=4, the gradient is 0, meaning that the tangent at that point is horizontal. 

c) find the full coordinate on the curve by finding f(4)

The function is $f(x)=\frac{1}{4}x^2-2x+3$f(x)=14x22x+3, so $f(4)=\frac{1}{4}4^2-2\times4+3=4-8+3=-1$f(4)=14422×4+3=48+3=1.  So the coordinate on the curve is $(4,-1)$(4,1)

d) use the point and the gradient to find the equation of the tangent.

The point is $(4,-1)$(4,1) and the gradient is $0$0.  The equation of the horizontal line is $y=-1$y=1

 

Worked Examples

Question 1

By considering the graph of $f\left(x\right)=-6$f(x)=6, find $f'\left(4\right)$f(4).

Question 2

By considering the graph of $f\left(x\right)=2x-3$f(x)=2x3, find $f'$f$\left(-4\right)$(4).

Question 3

Consider the parabola $f\left(x\right)=x^2+3x-10$f(x)=x2+3x10.

  1. Solve for the $x$x-intercepts. Write all solutions on the same line, separated by a comma.

  2. Determine the gradient of the tangent at the positive $x$x-intercept.

Question 4

Consider the tangent to the curve $f\left(x\right)=5\sqrt{x}$f(x)=5x at the point $\left(\frac{1}{9},\frac{5}{3}\right)$(19,53).

  1. Firstly, find the gradient of the function $f\left(x\right)=5\sqrt{x}$f(x)=5x at $x=\frac{1}{9}$x=19.

  2. Hence find the equation of the tangent to the curve $f\left(x\right)=5\sqrt{x}$f(x)=5x at the point $\left(\frac{1}{9},\frac{5}{3}\right)$(19,53).

    Express the equation of the tangent line in the form $y=mx+c$y=mx+c.

Question 5

Consider the tangent to the curve $f\left(x\right)=\frac{2}{x^3}$f(x)=2x3 at $x=-2$x=2.

  1. Firstly, find the gradient of the function $f\left(x\right)=\frac{2}{x^3}$f(x)=2x3 at $x=-2$x=2.

  2. Determine the $y$y-coordinate of the point on the tangent line.

  3. Hence find the equation of the tangent to the curve $f\left(x\right)=\frac{2}{x^3}$f(x)=2x3 at $x=-2$x=2.

    Express the equation of the tangent line in the form $y=mx+c$y=mx+c.

 

 

Outcomes

11.C.LD.1

Derivative introduced as rate of change both as that of distance function and geometrically, intuitive idea of limit. Definition of derivative, relate it to slope of tangent of the curve, derivative of sum, difference, product and quotient of functions. Derivatives of polynomial and trigonometric functions.

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