Find an equation for the tangent line to the graph of the function F(x)=√5x^2+1 at the point (4,9).

Answer 1

Finding the equation of the tangent line amounts to determining the linear function (y = mx + b) which passes through the point (4,9). The tangent line is a synonym for slope (the value of m in the linear function) but can also be equated to the derivative of a function evaluated at a particular point.

Given f(x) = √5*x² + 1

f'(x) = √5 * 2x = 2√5x --> based on the power-rule and associativity rule (sum of derivatives = derivative of sum)

f'(4) = 2√5*(4) = 8√5 = m

Using the point (4,9) to determine b from the linear equation now...

y = 8√5*x + b

9 = 8√5*(4) + b

b = 9 - 32√5

The tangent line passing through the point (4,9) is

y = 8√5 * x + ( 9 - 32√5 )

On the off chance the function is really intended to be f(x) = sqrt(5x^2 + 1) (note the parentheses used to ensure proper grouping), I will do this function as well...

f'(x) = 1/2 * 1/sqrt(5x^2 + 1) * 10x --> (power rule with chain rule)

f'(4) = 1/2 * 1/sqrt(81) * 10*4 = 1/2 * 40 * 1/9 = 20/9

y = (20/9)*x + b

9 = (20/9)*4 + b

b = 9 - 80/9 = 1/81

The tangent line passing through the point (4,9) is

y = (20/9) * x + 1/81

Answer 2

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