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Math  /  Calculus

Question5. Answer the following questions; a) Find the slope of the curve f(x)=x−1f(x)=\sqrt{x-1} at the point x=5x=5.
Ans: \qquad

Studdy Solution

STEP 1

1. The function given is f(x)=x−1f(x) = \sqrt{x-1}, and we need to find its slope at x=5x = 5.
2. The slope of the curve at a particular point is given by the derivative of the function evaluated at that point.
3. We need to find the derivative of f(x)f(x) and then evaluate it at x=5x = 5.

STEP 2

1. Find the derivative of the function f(x)=x−1f(x) = \sqrt{x-1}.
2. Evaluate the derivative at x=5x = 5 to find the slope of the curve at that point.

STEP 3

Rewrite f(x)=x−1f(x) = \sqrt{x-1} in a more convenient form for differentiation:
f(x)=(x−1)1/2 f(x) = (x-1)^{1/2}

STEP 4

Use the power rule for differentiation to find f′(x)f'(x):
f′(x)=ddx(x−1)1/2 f'(x) = \frac{d}{dx} (x-1)^{1/2}

STEP 5

Apply the chain rule to find the derivative. Let u=x−1u = x - 1, then f(x)=u1/2f(x) = u^{1/2}:
f′(x)=12(x−1)−1/2⋅ddx(x−1)=12(x−1)−1/2⋅1 f'(x) = \frac{1}{2} (x-1)^{-1/2} \cdot \frac{d}{dx}(x-1) = \frac{1}{2} (x-1)^{-1/2} \cdot 1

STEP 6

Simplify the expression for f′(x)f'(x):
f′(x)=12(x−1)−1/2=12x−1 f'(x) = \frac{1}{2} (x-1)^{-1/2} = \frac{1}{2\sqrt{x-1}}

STEP 7

Evaluate the derivative at x=5x = 5:
f′(5)=125−1=124=12⋅2=14 f'(5) = \frac{1}{2\sqrt{5-1}} = \frac{1}{2\sqrt{4}} = \frac{1}{2 \cdot 2} = \frac{1}{4}
The slope of the curve f(x)=x−1f(x) = \sqrt{x-1} at the point x=5x = 5 is 14\frac{1}{4}.

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