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Find $f(1)$ and $f'(1)$ if $2x+3y=5$ is the tangent of $f(x)$ at $x=1$.

Is this correct: $$2x+3y=5$$ $$3y=5-2x$$ $$y=5/3-(2/3)x$$ From here I get that $f'(1) = -2/3$. Here I am not sure how to continue for $f(1)$? If I just replace it in $y=5/3-2/3x$, I get $y=5/3-2/3=3/3=1$, but I am not sure if this is correct?

Alexey Burdin
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congruencer
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1 Answers1

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You are correct:

If $f$ is differentiable at $x_0$, the tangent line of $f$ at $x_0$ is by definition the line $y = f'(x_0) (x-x_0) + f(x_0) $. So if the line $y = \alpha x + \beta$ is the tangent line of $f$ at $x = x_0$ then

$$ f'(x_0) = \alpha $$

and

$$ f(x_0) = \alpha x_0 + \beta. $$

DodoDuQuercy
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