Suppose it is known that for a given differentiable function y=f(x), its tangent line (local linearization) at the point where a=−4 is given by t(x)=5−7(x+4). what must be the values of f(−4) and f′(−4)?

y(-4) = 5

y'(-4) = -7

Step-by-step explanation:

Hi!

Since the tangent line T and the curve y must coincide at x=-4

y(-4) = T(-4) = 5

On the other hand, the derivative of the curve evaluated at -4 y'(x=-4) must be the slope of the tangent line. Which inspecting the tangent line T(x) is -7

That is:

y'(-4) = -7

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step-by-step explanation:

it has changed from the width and the height.

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