Question

Determine the equation of the tangent to f(x) at x=5 ?

Grade 12 calculus

Answer 1

Please see below.

Explanation:

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If `f(x)=sqrt(2x-1`:

Slope of the tangent line to the curve can be calculated by taking the derivative of the function and evaluating it at the point of tangency:

`y=sqrt(2x-1)=(2x-1)^(1/2)`

`m=dy/dx=1/2(2x-1)^(-1/2)(2)=1/2(2(5)-1)^(-1/2)(2)=1/2*9^(-1/2)(2)=1/9^(1/2)=1/sqrt9=1/(+-3)=1/3, -1/3`

Because the function can only be positive,

The equation of the tangent line is:

`y=mx+b`

`y=1/3x+b`

We use the coordinates of the point of tangency to solve for `b`:

`x=5, :. y=sqrt(2(5)-1)=sqrt9=3`

`3=1/3(5)+b`

`9=5+3b`

`3b=4`

`b=4/3`

The equation of the tangent line is:

`y=1/3x+4/3`

Here is the graph of the function and its tangent:

If `f(x)=sqrt(2x)-1`:

`y=sqrt(2x)-1=sqrt2x^(1/2)-1`

`m=dy/dx=sqrt2*1/2x^(-1/2)=sqrt2*1/2(5)^(-1/2)=sqrt2/(2sqrt5)=(sqrt2sqrt5)/10=sqrt10/10`

`x=5, :. y=sqrt(2(5))-1=sqrt10-1`

`y=mx+b`

`y=sqrt10/10x+b`

`sqrt10-1=sqrt10/10(5)+b`

`sqrt10-1=sqrt10/2+b`

`2sqrt10-2=sqrt10+2b`

`sqrt10-2=2b`

`b=(sqrt10-2)/2`

Equation of the tangent line is:

`y=sqrt10/10x+(sqrt10-2)/2`

Here is the graph of the function and its tangent: