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Calculus 1 : How to find slope by graphing functions

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Example Question #1 :Slope

What is the slope of the tangent line of f(x) = 3x⁴– 5x³– 4x at x = 40?

Possible Answers:

768,000

None of the other answers

331,841

684,910

743,996

Correct answer:

743,996

Explanation:

The first derivative is easy:

f'(x) = 12x³– 15x²– 4

The slope of the tangent line is found by calculating f'(40) = 12 * 40³– 15 * 40²– 4 = 768,000 – 24,000 – 4 = 743,996

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Example Question #2 :Slope

Find the slope of the line tangent to y(b) whenis equal to.

y(b)=b^7-b^5+4b^2

Possible Answers:

3216

320160

3160

Correct answer:

320160

Explanation:

To find the slope of a tangent line, we need to find the first derivative of the function at that point. In other words, we need y'(6).

y(b)=b^7-b^5+4b^2

Taking the first derivative using the Power Rule $f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ we get the following.

y'(b)=7b^6-5b^4+8b

Substituting in 6 for b and solving we get:

$y'(6)=7\cdot6^6-5\cdot6^4+8\cdot6=320160$ .

So our answer is 320160

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Example Question #3 :Slope

Find function which gives the slope of the line tangent to j(x) .

$j(x)=\frac{4x^5}{5}-x^4+x^2$

Possible Answers:

16x^3-12x^2+2

x^4-x^3+x

4x^4-4x^3+2

3x^4-12x^3+2x

4x^4-4x^3+2x

Correct answer:

4x^4-4x^3+2x

Explanation:

找到一个切线的斜率,我们需要first derivative.

Recall that to find the first derivative of a polynomial, we need to decrease each exponent by one and multiply by the original number.

$j(x)=\frac{4x^5}{5}-x^4+x^2$

j'(x)=4x^4-4x^3+2x

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Example Question #1 :Slope

Find the slope of the line tangent to b(s) at s=16 .

b(s)=14s^2+6s-4

Possible Answers:

545

144

454

544

343

Correct answer:

454

Explanation:

The slope of the tangent line can be found easily via derivatives. To find the slope of the tangent line at s=16, find b'(16) using the power rule on each term which states:

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$

Applying this rule we get:

b(s)=14s^2+6s-4

b'(s)=28s+6

$b'(16)=28\cdot 16+6=454$

Therefore, the slope we are looking for is 454.

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Example Question #2 :Slope

Find the slope of f(x) at $x=\pi$ .

f(x)=ln(x^2 +2)-cos(17x)

Possible Answers:

17.529

0.529

0.629

10.229

Correct answer:

0.529

Explanation:

To find the slope of the line at that point, find the derivative of f(x) and plug in that point.

$f'(x)= \frac{2x}{x^2 +2}+17sin(17x)$

Remember that the derivative of cos(bx) = -bsin(bx) and the derivative of $ln(u) = \frac{u'}{u}$

Now plug in $\pi$

$f'(\pi )=\frac{2\pi}{\pi ^2 +2}=.529$

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Example Question #6 :Slope

Find the slope of f(x) at x=2 given f''(x) . Assume the integration constant is zero.

f''(x)= x+2

Possible Answers:

Correct answer:

Explanation:

The first step here is to integrate f''(x) in order to get f'(x) .

$\int f''(x)dx =\int(x+2)dx=\frac{x^2}{2} +2x +C$

Here the problem tells us that the integration constant C=0 , so

$f'(x)=\frac{x^2}{2} + 2x$

Plug in x=2 here

f'(2)=6

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Example Question #7 :Slope

Consider the curve

y=ln[x]+3x^2 .

What is the slope of this curve at x=1.5 ?

Possible Answers:

$\frac{17}{3}$

$\frac{29}{3}$

$\frac{34}{3}$

Correct answer:

$\frac{29}{3}$

Explanation:

The slope of a curve at any point is equal to the derivative of the curve at that point.

Remembering that the derivative of $ln(x)\rightarrow \frac{1}{x}$ and using the power rule on the second term we find the derivative to be:

$\frac{dy}{dx}=\frac{1}{x}+6x$ .

Pluggin in x=1.5 we find that the slope is $\frac{29}{3}$ .

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Example Question #8 :Slope

Find the line tangent to $y=e^{5x}$ at x=3 .

Possible Answers:

$y=5e^{15}x-14e^{15}$

$y=5e^{15}x-3$

$y=e^{15}x+e^{15}$

$y=5e^{15}x-14e^{15}$

$y=5e^{15}x-2e^{15}$

Correct answer:

$y=5e^{15}x-14e^{15}$

Explanation:

Find the line tangent to $y=e^{5x}$ at x=3 .

First, we find:

$y=e^{5x}=e^{5\cdot3}=e^{15}$

Next, we find the derivative:

$\frac{dy}{dx}=5e^{5x}$

Therefore, the slope at x=3 is:

$\frac{dy}{dx}=5e^{5x}=5e^{5\cdot3}=5e^{15}$ .

Using point-slope form, we can write the tangent line:

$y=5e^{15}(x-3)+e^{15}$

Simplifying this gives us:

$y=5e^{15}x-14e^{15}$

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Example Question #9 :Slope

An isosceles triangle has one point at (6,2) , one point at (6,10) and one point on the-axis. What is the slope of the line between the point on the-axis and (6,10) ?

Possible Answers:

5/3

3/2

1/2

2/3

Correct answer:

2/3

Explanation:

The other point of the triangle must be at (0,6) as it must be equidistant from the other two points of the triangle. Since all points on the y-axis areunits away from the other points in thedirection, the third point must be equidistant in thedirection from both (6,2) and (6,10) . The distance between these points is, so the third point must have a y-value of 2+(8/2)=6 . The third point is now at (0,6) so the slope of the line from (0,6) to (6,10) is as follows.

$\\m=\frac{y_2-y_1}{x_2-x_1}\\ \\=\frac{10-6}{6-0}\\ \\ =\frac{4}{6}=\frac{2}{3}$

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Example Question #10 :Slope

What is the slope of the line tangent to the graph of $f(x)=\sin(2x)$ at $x=\pi/2$ ?

Possible Answers:

1/2

-1/2

Correct answer:

Explanation:

We must take the derivative of the function using the chain rule yielding $f'(x)=2\cos(2x)$ .

The chain rule is $[f(g(x))]'=f'(g(x))\cdot g'(x)$ .

Also remember that the derivative of sin(x) is cos(x) .

Applying these rules we get the following.

$\\f(x)=\sin(2x)\\ \\ u=2x\\ \\ \frac{du}{dx}=2\\ \\ \frac{d}{dx}\sin(u)=\cos(u)\frac{du}{dx}=2\cos(u)\\ \\ f'(x)=2\cos(2x)$

Plugging in the value forwe get $2\cos(\pi)$ which is.

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Calculus 1 : How to find slope by graphing functions

Study concepts, example questions & explanations for Calculus 1

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