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I am being asked to find the slope of a tangent to a curve when $x=4$.

The equation I have is $f(x) = 4x^3 - 5x + 2\sqrt{x}$

I'm a beginner and I must say that I'm having a hard time with this. I have read this previous post but I'm not sure how to apply it to my problem.

Thanks

Community
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user142050
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  • The slope of the tangent to a (differentiable) curve at a certain point $x=x_0$ is $f'(x_0)$, the derivative of the function at that point. – fixedp Apr 09 '14 at 20:27
  • @user142050 That previous post isn't going to help a whole lot. Do you know how to take the derivative of your function? – imranfat Apr 09 '14 at 20:28

3 Answers3

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$$f(x)=4x^3-5x+2\sqrt{x}\\ \implies f^\prime(x)=12x^2-5+x^{-1/2}\\ \implies f^\prime(4)=12\times16-5+\dfrac{1}{2}=187.5\\ \implies y-f(4)=f^\prime(4)(x-4) \text{Now, }f(4)=4\times64-5\times4+2\times2=240\\ \implies \text{Tangent line: }y-240=187.5(x-4)$$

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Note that $$f(x) = 4x^3 - 5x + 2\sqrt{x} = 4x^3 - 5x + 2x^{1/2}.$$

What does $f'(x)$ represent, and in particular $f'(4)$? Can you compute it?

gt6989b
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The slope tangent to $f$ at $x = 4 $ is the derivative of $f$ evaluated at $x = 4 $: hence using power rule, we have

$$f'(x) = 12 x^2 - 5 + \frac{1}{\sqrt{x} } $$

$$ \therefore f'(4) = 12(4)^2 - 5 + \frac{1}{\sqrt{4}} = 192 - 5 + \frac{1}{2} = 187 + \frac{1}{2} = 187.5 $$