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If f (x)=√1+8x, find ƒ '(a) for a = 3 and the equation of the tangent line to f(x) at x=a.

a. f'(a) = 5/4 ; equation of the tangent line is y = 5x-13/4
b. f'(a) = -4/5 : equation of the tangent line is y = - 4x + 13/5
c. f'(a) = 4/5 : equation of the tangent line is y = 4x + 13/5
d. f'(a) = -5/4 ; equation of the tangent line is y = -5x-13/4

1 Answer

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Final answer:

To find f'(a), differentiate f(x) with respect to x and evaluate the derivative at x=a. Then, use the point-slope form of a line to find the equation of the tangent line.

Step-by-step explanation:

To find f'(a), we need to differentiate f(x) with respect to x and then evaluate the derivative at x=a. Let's start by differentiating f(x) = √(1+8x) using the chain rule.

f '(x) = 1/2(1+8x)^(-1/2) * 8

Simplifying this, we get f '(x) = 4/(1+8x)^(1/2)

To find f '(a), we substitute x=a into the derivative expression: f '(a) = 4/(1+8a)^(1/2)

Now let's find the equation of the tangent line to f(x) at x=a. The equation of a tangent line is in the form y = mx + c, where m is the slope of the line and c is the y-intercept. The slope of the tangent line is given by f '(a). We can use the point-slope form of a line to find the equation.

y - f(a) = f '(a) * (x - a)

Substituting f(x) = √(1+8x) and x=a, we get y - f(a) = f '(a) * (x - a).

Therefore answer is c. f'(a) = 4/5 : equation of the tangent line is y = 4x + 13/5.

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