← Back to learning objective

Estimating tangent slope from secant slopes

Consider the function f(x)=xf(x) = \sqrt{x}. We want to estimate the slope of the tangent line to the graph of ff at x=4x = 4 by computing the slopes of secant lines through the points (4,f(4))(4, f(4)) and (4+h,f(4+h))(4+h, f(4+h)) for smaller and smaller values of hh.

The slope of the secant line through (4,f(4))(4, f(4)) and (4+h,f(4+h))(4+h, f(4+h)) is given by

f(4+h)−f(4)h.\frac{f(4+h) - f(4)}{h}.

Compute this secant slope for h=0.1h = 0.1, h=0.01h = 0.01, and h=0.001h = 0.001 (use these as scratch work). Based on the trend in your computed values, what is your best estimate for the slope of the tangent line to f(x)=xf(x) = \sqrt{x} at x=4x = 4?

Hint

Compute 4+h\sqrt{4+h} for each value of hh, subtract 4=2\sqrt{4} = 2, and divide by hh. Watch how the resulting numbers change as hh shrinks — where do they appear to be heading?

Step-by-step solution

For each hh, compute 4+h−2h\dfrac{\sqrt{4+h} - 2}{h}. For h=0.1h = 0.1: 4.1≈2.02485\sqrt{4.1} \approx 2.02485, giving slope ≈0.024850.1≈0.2485\approx \dfrac{0.02485}{0.1} \approx 0.2485. For h=0.01h = 0.01: 4.01≈2.0024984\sqrt{4.01} \approx 2.0024984, giving slope ≈0.00249840.01≈0.24984\approx \dfrac{0.0024984}{0.01} \approx 0.24984. For h=0.001h = 0.001: 4.001≈2.00024998\sqrt{4.001} \approx 2.00024998, giving slope ≈0.000249980.001≈0.24998\approx \dfrac{0.00024998}{0.001} \approx 0.24998.

As hh gets smaller (0.1, 0.01, 0.001), the secant slopes get closer and closer to 0.250.25, approaching it from below. This trend strongly suggests that the tangent line slope at x=4x=4 is 14=0.25\dfrac{1}{4} = 0.25, consistent with the exact derivative f′(x)=12xf'(x) = \dfrac{1}{2\sqrt{x}} evaluated at x=4x=4.

Related problems

Practice similar Calculus problems interactively.

Start Closing the Gap