Question

Determine the value or values of x for which the tangent to f is horizontal by first finding the derivative of f with respect to x then solving f'(x) = 0 for x. PART 1. f(x) = x^2 / (x + 9) f'(x) = NOTE: for this problem, you should use the definition of derivative, f'(x) = lim_{h->0} (f(x + h) - f(x)) / h or the equivalent form f'(x) = lim_{z->x} (f(z) - f(x)) / (z - x) PART 2. The tangent to the curve is horizontal at:

          Determine the value or values of x for which the tangent to f is horizontal by first finding the derivative of f with respect to x then solving f'(x) = 0 for x.

PART 1.
f(x) = x^2 / (x + 9)
f'(x) =

NOTE: for this problem, you should use the definition of derivative, f'(x) = lim_{h->0} (f(x + h) - f(x)) / h or the equivalent form f'(x) = lim_{z->x} (f(z) - f(x)) / (z - x)

PART 2.
The tangent to the curve is horizontal at:

Determine the value or values of x for which the tangent to f is horizontal by first finding the derivative of f with respect to x then solving f'(x) = 0 for x.

PART 1.
f(x) = x^2 / (x + 9)
f'(x) =

NOTE: for this problem, you should use the definition of derivative, f'(x) = limh->0 (f(x + h) - f(x)) / h or the equivalent form f'(x) = limz->x (f(z) - f(x)) / (z - x)

PART 2.
The tangent to the curve is horizontal at:

Added by Kristina S.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

10/11/2022

Step 1

Step 1:** Find the derivative of \( f(x) = x^2 + 9 \) with respect to \( x \): \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] \[ = \lim_{h \to 0} \frac{(x+h)^2 + 9 - (x^2 + 9)}{h} \] \[ = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 + 9 - x^2 - 9}{h} \] \[ = \lim_{h Show more…

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Determine the value or values of x for which the tangent to f is horizontal by first finding the derivative of f with respect to x then solving f'(x) = 0 for x. NOTE: for this problem, you should use the definition of derivative, f'(x) = lim_{h->0} (f(x+h) - f(x))/h or the equivalent form f'(x) = lim_{z->x} (f(z) - f(x))/(z - x). PART 2. The tangent to the curve is horizontal at: