If f(x) is differentiable for the closed interval [-1, 4] such that f(-1) = -3 and f(4) = 12, then there exists a value c, -1< c < 4 such that

f '(c) = 3

f '(c) = 0

f(c) = -15

f (c) = 3

Respuesta :

Answer: Choice A,   f ' (c) = 3

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Work Shown:

f(-1) = -3 means the point (-1,-3) is on the f(x) curve.

f(4) = 12 means (4,12) is on the f(x) curve as well.

Compute the slope of the line through those two points.

Slope formula

m = (y2 - y1)/(x2 - x1)

m = (12 - (-3))/(4 - (-1))

m = (12+3)/(4+1)

m = 15/5

m = 3

The slope of the secant line through (-1,-3) and (4,12) is m = 3

Through the mean value theorem (MVT), there exists at least one value c such that f ' (c) = 3, where -1 < c < 4, and f(x) is a continuous and differentiable function on this interval in question.

Visually, there exists at least one tangent line that has the same slope of the secant line mentioned. Lines with equal slopes, and different y intercepts, are parallel.

Answer: The answer is A

(7x3=21)- 10= 11 - 11=0

Step-by-step explanation: