For f(x)=√(5x+10), find f′(x) using the definition f′(x)= lim, h→0, {f(x+h)−f(x)} / h . Then, using this, find the tangent line to the graph of y= √(5x+10) at x=3

Question

For f(x)=√(5x+10), find f′(x) using the definition f′(x)= lim, h→0, {f(x+h)−f(x)} / h . Then, using this, find the tangent line to the graph of y= √(5x+10) at x=3. Enter all values as integers, or fractions in lowest terms.

Answer 1

f(x) = √(5x+10)

f(x+h) = √[5(x+h)+10] = √(5x+5h+10)

So [f(x+h) - f(x)]/h = [√(5x+5h+10) - √(5x+10)]/h

Now multiply numerator and denominator by the conjugate of the numerator:

The numerator then becomes: 5x + 5h + 10 - (5x + 10) = 5x + 5h + 10 - 5x - 10 = 5h

Then the "h" in the numerator cancels with the "h" in the denominator resulting in:

lim(h→0) 5/[√(5x+5h+10) + √(5X + 10)]

Which is 5/(√(5x + 10) + √(5x + 10) = 5/(2√5x+10)

So f '(x) = 5/(2√5x+10)

At x = 3, the slope of the tangent line is then: f ' (3) = 5/2√5(3)+10) = 5/(2√25) = 5/10 = 1/2

Use that in the point-slope form of a line: y - y₁ = m(x - x₁). So m = 1/2. To get (x₁, y₁), plug in x = 3 into the original function. Then plug in the point.